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INDUSTRIAL  SERIES 

SHOP  MATHEMATICS 

PART  I 

SHOP  ARITHMETIC 

PREPARED  IN  THE 

EXTENSION  DIVISION  OF 
THE  UNIVERSITY  OF  WISCONSIN 

BY 
EARLE  B.  NORRIS,  M.  E. 

ASSOCIATE    PROFESSOR   OF  MECHANICAL  ENGINEERING,  IN  CHARGE 

OF  MECHANICAL  ENGINEERING   COURSES   IX   THE 

UNIVERSITY  EXTENSION   DIVISION 

AND 

KENNETH  G.  SMITH,  A.  B.,  B.  S. 

ASSOCIATE   PROFESSOR  OF  MECHANICAL  ENGINEERING,   DISTRICT 

REPRESENTATIVE    IN   CHARGE   OF   FIRST   DISTRICT,   THE 

UNIVERSITY   EXTENSION    DIVISION,  MILWAUKEE 


FIRST  EDITION 
SECOND  IMPRESSION 


McGRAW-HILL   BOOK  COMPANY 

239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.  C. 

1912 


$883 


COPYRIGHT,  1912,  BY  THE 

BOOK  COMPANY 


7113 


THE. MAPLE. PRESS- YORK. PA 


N  n 

\r 


PREFACE 

The  aim  of  this  book  is  to  teach  the  fundamental  principles  of 
mathematics  to  shop  men,  using  familiar  terms  and  processes, 
and  giving  such  applications  to  shop  problems  as  will  maintain 
the  interest  of  the  student  and  develop  in  him  an  ability  to  apply 
the  mathematical  and  scientific  principles  to  his  every  day 
problems  of  the  shop.  The  problems  and  applications  relate 
largely  to  the  metal  working  trades.  It  has,  however,  been  the 
aim  in  preparing  this  volume  not  to  apply  the  work  to  these 
particular  trades  so  closely  but  that  it  shall  be  of  interest  and 
value  to  men  in  other  lines  of  industry. 

This  volume  presents  the  first  half  of  the  instruction  papers  in 
Shop  Mathematics  as  developed  and  used  by  the  Extension 
Division  of  the  University  of  Wisconsin.  As  here  offered,  it 
embodies  the  point  of  view  obtained  through  apprenticeship 
and  shop  experience  as  well  as  the  experience  gained  through  its 
use  during  the  past  four  years  as  a  text  for  both  correspondence 
and  class  room  instruction.  It  is  believed  that  the  book  will 
be  found  suitable  for  home  study  and  for  use  as  a  text  in  trade, 
industrial,  and  continuation  schools. 

The  instruction  in  arithmetic  ends  with  Chapter  XII.  The 
remaining  chapters  are  introduced  to  give  further  practice  in 
calculation  and  to  develop  an  ability  to  handle  simple  formulas, 
as  well  as  to  impart  a  knowledge  of  the  principles  of  machines. 
The  second  volume  will  take  up  more  fully  the  use  of  formulas 
and  will  teach  the  principles  of  geometry  and  trigonometry  as 
applied  to  shop  work. 

The  authors  are  indebted  to  Mr.  F.  D.  Crawshaw,  Professor  of 
Manual  Arts  in  The  University  of  Wisconsin,  for  a  careful 
reading  of  the  proof  and  for  valuable  criticisms  and  suggestions. 

E.  B.  N. 

THE  UNIVERSITY  OF  WISCONSIN, 

MADISON,  Wis. 

June  1,  1912. 


CONTENTS 

CHAPTER  I 

COMMON  FRACTIONS 
ART.  PAGE 

1.  Why  we  use  fractions 1 

2.  Definition  of  a  fraction      2 

3.  The  denominator 2 

4.  The  numerator 2 

5.  Writing  and  reading  fractions      2 

6.  Proper  fractions      3 

7.  Improper  fractions 3 

8.  Mixed  numbers 3 

9.  Reduction  of  fractions 3 

10.  Reduction  to  higher  terms 4 

11.  Reduction  to  lower  terms 4 

12.  Reduction  of  improper  fractions 5 

13.  Reduction  of  mixed  numbers 6 

CHAPTER   II 

ADDITION  AND  SUBTRACTION  OF  FRACTIONS 

14.  Common  denominators 9 

15.  To  find  the  L.  C.  D 10 

16.  To  reduce  to  the  L.  C.  D 10 

17.  Addition  of  fractions 11 

18.  Subtraction  of  fractions 12 

CHAPTER  III 
MULTIPLICATION  AND  DIVISION  OF  FRACTIONS 

19.  A  whole  number  times  a  fraction 17 

20.  "Of"  means  times 17 

21.  A  fraction  times  a  fraction 18 

22.  Multiplying  mixed  numbers 18 

23.  Cancellation 19 

24.  Division — the  reverse  of  multiplication      21 

25.  Compound  fractions 21 

26.  How  to  analyze  practical  problems 22 

CHAPTER  IV 
MONEY  AND  WAGES 

27.  U.  S.  money 24 

28.  Addition 25 

• 

vii 


viii  CONTENTS 

ART.  PAOE 

29.  Subtraction '. 26 

30.  Multiplication , 26 

31.  Division •    •. 26 

32.  Reducing  dollars  to  cents '27 

33.  Reducing  cents  to  dollars 28 

34.  The  mill 28 

35.  Wage  calculations      29 

CHAPTER  V 
DECIMAL  FRACTIONS 

36.  What  are  decimals? 33 

37.  Addition  and   subtraction 35 

38.  Multiplication 36 

39.  Short  cuts 37 

40.  Division 37 

41.  Reducing  common  fractions  to  decimals 39 

42.  Complex  decimals 40 

43.  The  micrometer      40 

CHAPTER  VI 

PEKCENTAGE 

44.  Explanation 44 

45.  The  uses  of  percentage      46 

46.  Efficiencies 47 

47.  Discount      . .' 47 

48.  Classes  of  problems 48 

CHAPTER  VII 

CIRCUMFERENCES  OF  CIRCLES:  CUTTING  AND  GRINDING  SPEEDS 

49.  Shop  uses 51 

50.  Circles 51 

51.  Formulas 52 

52.  Circumferential  speeds 54 

53.  Grindstones  and  emery  wheels 55 

54.  Cutting  speeds 57 

55.  Pulleys  and  belts 58 

CHAPTER  VIII 
RATIO  AND  PROPORTION 

56.  Ratios 59 

57.  Proportion '.'...  60 

58.  Speeds  and  diameters  of  pulleys 63 

59.  Gear  ratios 64  ' 


CONTENTS  ix 

CHAPTER  IX 
PULLEY  AND  GEAR  TRAINS;  CHANGE  GEARS 

Am.  J'ACib; 

60.  Direct  and  inverse  proportions 65 

61.  Gear  trains 60 

62.  Compound  gear  and  pulley  trains 68 

63.  Screw  cutting 72 

CHAPTER  X 
AREAS  AND  VOLUMES  OF  SIMPLE  FIGURES 

64.  Squares 75 

65.  Square  root 75 

66.  Cubes  and  higher  powers 75 

67.  Square  measure      76 

68.  Area  of  a  circle 77 

69.  The  rectangle      79 

70.  The  cube 80 

71.  Volumes  of  straight  bars 80 

72.  Weights  of  metals      82 

73.  Short  rule  for  plates 83 

74.  Weight  of  casting  from  pattern 83 

CHAPTER  XI 
SQUARE  ROOT 

75.  The  meaning  of  square  root 85 

76.  Extracting  the  square  root 86 

77.  Square  roots  of  mixed  numbers 87 

78.  Square  roots  of  decimals 87 

79.  Rules  for  square  root 88 

80.  The  law   of  right  triangles 89 

81.  Dimensions  of  squares  and  circles 91 

82.  Dimensions  of  rectangles 91 

83.  Cube  root 92 

CHAPTER  XII 
MATHEMATICAL  TABLES  (CIRCLES,  POWERS,  AND  ROOTS) 

84.  The  value  of  tables       94 

85.  Explanation  of  the  tables 95 

86.  Interpolation 96 

87.  Roots  of  numbers  greater  than  1000 5)7 

88.  Cube  roots  of  decimals      ! 98 

89.  Square  root  by  the  table 99 


x  CONTENTS 

CHAPTER  XIII 

LEVERS 

ART.  I'A<;K 

90.  Types  of  machines 115 

91.  The  lever 115 

92.  Three  classes  of  levers 117 

93.  Compound  levers 118 

94.  Mechanical  Advantage 119 

95.  The  wheel  and  axle 120 

CHAPTER  XIV 

TACKLE  BLOCKS 

96.  Types  of  blocks  .    .    .    , 123 

97.  Differential  pulleys 126 

CHAPTER  XV 
THE  INCLINED  PLANE  AND  SCREW 

98.  The  use  of  inclined  planes 130 

99.  Theory  of  the  inclined  plane 130 

100.  The  wedge 132 

101.  The  jack  screw 133 

102.  Efficiencies 134 

CHAPTER  XVI 
WORK  POWER  AND  ENERGY;  HORSE-POWER  OF  BELTING 

103.  Work 137 

104.  Unit  of  work  .    .    . 137 

105.  Power 138 

106.  Horse-power  of  belting 139 

107.  Widths  of  belts 140 

108.  Rules  for  belting 141 

CHAPTER  XVII 
HORSE-POWER  or  ENGINES 

109.  Steam  engines 145 

110.  Gas  engines 147 

111.  Air  compressors      149 

112.  Brake  horse-power 11!) 

113.  Frictional  horsc-pbwer 151 

114.  Mechanical  efficiency 151 


CONTENTS  xi 

CHAPTER  XVIII 

MECHANICS  OF  FLUIDS 
ART.  PAGE 

115.  Fluids 153 

116.  Specific  gravity 153 

117.  Transmission  of  pressure  through  fluids 154 

118.  The  hydraulic  jack 155 

119.  Hydraulic  machinery 158 

120.  Hydraulic  heads 158 

121.  Steam  and  air • 159 

CHAPTER  XIX 
HEAT 

122.  Nature  of  heat 163 

123.  Temperatures      164 

124.  Expansion  and  contraction 168 

125.  Allowances  for  shrink  fits 171 

HAPTER  XX 
STRENGTH  OF  MATERIALS 

126.  Stresses 173 

127.  Ultimate  stresses 174 

128.  Safe  working  stresses 174 

129.  Strengths  of  bolts 175 

130.  Strengths  of  hemp  ropes 177 

131.  Wire  ropes  and  cables 177 

132.  Strengths  of  chains 177 

133.  Columns  .  .178 


SHOP   ARITHMETIC 


CHAPTER  I 


COMMOX  FRACTIONS 

1.  Why  We  Use  Fractions.  —  When  we  find  it  necessary  to  deal 
with  things  that  are  less  than  one  unit,  we  must  use  fractions. 
A  machinist  cannot  do  all  his  work  in  full  inches  because  it  is 
generally  impossible  to  have  all  measurements  in  exact  inches. 
Consequently,  for  measurements  less  than  1  in.,  he  uses  fractions 
of  an  inch;  he  also  makes  use  of  fractions  for  measurements 
between  one  whole  number  of  inches  and  the  next  whole  number. 
If  a  bolt  is  wanted  longer  than  4  in.  but  shorter  than  5  in.,  it 
would  be  4  in.  and  a  fraction  of  an  inch.  This  fraction  of  an 
inch  might  be  nearly  a  whole  inch  or  it  might  be  a  very  small  part 
of  an  inch.  The  system  used  to  designate  parts  of  a  unit  is 


FIG.  1. 

easily  seen  by  looking  at  a  machinist's  scale  or  at  a  foot-rule  of 
any  sort.  Each  inch  on  the  scale  is  divided  into  a  number  of 
equal  parts.  A  wooden  foot-rule  usually  has  eight  or  sixteen 
parts  to  each  inch,  while  a  machinist's  steel  scale  has  much, finer 
divisions.  Now,  if  we  want  to  measure  a  piece  of  steel  which  is 
not  an  inch  long,  we  hold  a  scale  against  it,  as  dn  Fig.  1,  and  find 
out  how  many  of  these  divisions  of  an  inch  it  takes  to  equal  the 
length  of  the  piece.  The  scale  in  Fig.  1  is  3  in.  long  and  each 
inch  is  divided  into  eight  parts.  We  see  that  this  piece  'is  as 

1 


2  SHOP  ARITHMETIC 

long  as  five  of  these  eight  parts  of  an  inch,  or  we  say  that  it  is 
"five-eighths  "  of  an  inch  long. 

2.  Definition  of  a  Fraction. — A  Fraction  is  one  or  more  of  the 
equal  parts  into  which  anything  may  be  divided.     Every  fraction 
must  contain  two  numbers,  a  numerator  and  a  denominator. 
These  are  called  the  terms  of  a  fraction. 

3.  The  Denominator. — The  Denominator  tells  into  how  many 
equal  parts  the  unit  is  divided.     In  the  case  shown  in  Fig.  1, 
1  in.  was  the  unit  and  it  was  divided  into  eight  equal  parts. 
The  denominator  in  this  case  was  eight. 

4.  The  Numerator. — The  Numerator  shows  how  many  of  these 
parts  are  taken.     In  giving  the  length  of  the  piece  of  steel  in 
Fig.  1,  we  divided  the  inch  into  eight  parts  and  took  five   of 
them  for  the  length.     Five  is  the  numerator  and  eight  is  the 
denominator. 

5.  Writing  and  Reading  Fractions. — In  writing  fractions,  the 
numerator  is  placed  over  the  denominator  and  either  a  slanting 
line,  as  in  5/8,  or  a  horizontal  line,  as  in  f ,  drawn  between  them. 
The  horizontal  line  is  the  better  form  to  use,  as  mistakes  are 
easily  made  when  a  whole  number  and  a  fraction  with  a  slanting 
line  are  written  close  together. 

-j  is  read  one-fourth  or  one-quarter. 

-=  is  read  one-half. 

2  is  read  three-fourths  or  three-quarters. 

-=  is  read  five-eighths. 
o 

3 

=  is  read  three-sevenths. 

We  can  have  fractions  of  all  sorts  of  things  besides  inches. 
An  hour  of  time  is  divided  into  sixty  equal  parts  called  minutes. 
A  minute  is  merely  -fa  of  an  hour.  Likewise,  20  minutes  is  f  $ 
of  an  hour.  In  the  same  way,  1  second  is  -^  of  a  minute. 

In  the  early  days,  before  we  had  the  unit  called  the  inch,  the 
foot  was  the  common  unit  for  measuring  lengths.  When  it  was 
necessary  to  measure  lengths  less  than  1  ft.,  fractions  of  a  foot 


COMMON  FRACTIONS  3 

were  used.  This  got  to  be  too  troublesome,  so  one-twelfth  of  a 
foot  was  given  the  name  of  inch  to  avoid  using  so  many 
fractions.  For  instance,  where  formerly  one  said  f\  of  a  foot, 
we  can  now  say  5  in.  This  shows  how  the  use  of  a  smaller  unit 
reduces  the  use  of  fractions.  In  Europe,  a  unit  called  the 
millimeter  is  used  in  nearly  all  shop  work.  This  is  so  small,  being 
only  about  T£T  of  an  inch,  that  it  is  seldom  necessary  in  shop 
work  to  use  fractions  of  a  millimeter. 

6.  Proper  Fractions. — If  the  numerator  and  denominator  of  a 
fraction  are  equal,  the  value  of  the  fraction  is  1,  because  there 
are  just  as  many  parts  taken  as  there  are  parts  in  one  unit. 

4_!      8  10 

4~        8~        10" 

In  each  of  these  cases,  the  numerator  shows  that  we  nave  taken 
the  full  number  of  parts  into  which  the  unit  has  been  divided. 
Consequently,  each  of  the  fractions  equals  a  full  unit,  or  1. 

A  Proper  Fraction  is  one  whose  numerator  is  less  than  the 
denominator.  The  value  of  a  proper  fraction,  therefore,  is 
always  less  than  1. 

3  5    7  27 

A'  T«'  Q'  QO  are  a^  Pr°Per  fractions. 

4  ID  o  ou 

7.  Improper  Fractions. — An  Improper  Fraction  is  one  whose 
numerator  is  equal  to  or  larger  than  the  denominator.     There- 
fore, an  improper  fraction  is  equal  to,  or  more  than  1. 

24   14   17  64 

To'  IP  T«'  «7  are  a^  improper  fractions. 

1  _.     o      It)    C)4 

8.  Mixed  Numbers. — A  Mixed  Number  is  a  whole  number  and 
a  fraction  written  together:  for  example,  4£  is  a  mixed  number. 
4£  is  read  four  and  one-half  and  means  four  whole  units  and 
one-half  a  unit  more. 

9.  Reduction  of  Fractions. — Quite  often  we  find  it  desirable  to 
change  the  form  of  a  fraction  in  order  to  make  certain  calcula- 
tions; but,  of  course,  the  real  value  of  the  fraction  must  not  be 
changed.     The  operation  of  changing  a  fraction  from  one  form 
to  another  without  changing  its  value  is  called  Reduction. 

By  referring  to  the  scale  in  Fig.  2  it  will  be  seen  that,  if  we 
take  the  first  inch  and  divide  it  into  8  parts,  each  \  in.  will  con- 


SHOP  ARITHMETIC 


tain  4  of  these  parts.  Hence,  ^  in.  =  £  in.  In  this  case,  we 
make  the  denominator  of  the  fraction  4  times  as  large,  by  making 
4  times  as  many  parts  in  the  whole.  It  then  takes  a  numerator 
4  times  as  large  to  represent  the  same  fractional  part  of  an  inch. 
This  relation  holds  whether  we  are  dealing  with  inches  or  with 
any  other  thing  as  a  unit. 


FIG.  2. 

10.  Reduction  to  Higher  Terms. — When  we  raise  a  fraction  to 
higher  terms,  we  increase  the  number  of  parts  in  the  whole,  as 
just  shown,  and  this  likewise  increases  the  number  of  parts 
taken.  Therefore,  the  numerator  and  denominator  both  become 
larger  numbers. 


1  .  4. 

-  m.  =     in. 


A 

16 


10. 
=  32m' 


A  fraction  is  raised  to  higher  terms  by  multiplying  both  numer- 
ator and  denominator  by  the  same  number. 

Examples : 


1X4 


Similarly, 

-!=  5X2  =10 

16     16X2     32 
2 

Suppose  we  want  to  change  y^  of  an  inch  to  64ths.  To  get  64  for  the 
denominator,  we  must  multiply  16  by  4  and,  therefore,  must  multiply  3  by 
the  same  number. 

3  =  3X4  ^12 
lo  =  16X4     64 

11.  Reduction  to  Lower  Terms. — When  we  reduce  a  fraction 
to  lower  terms,  we  reduce  the  number  of  parts  into  which  the 
whole  unit  is  divided.  This  likewise  reduces  the  number  of 
parts  which  are  taken. 

4  .         1  .  2   .         1  . 


A  fraction  is  reduced  to  lower  terms  by  dividing  both  numer- 
ator and  denominator  by  the  same  number.     When  there  is  no 


COMMON  FRACTIONS  5 

number  -which  will  exactly  divide  both  numerator  and  denomina- 
tor, the  fraction  is  already  in  its  lowest  terms. 

Example : 

Reduce  ._  to  its  lowest  terms. 

36       36  4-2^18-4-2^  9 
128     128-5-2     64-J-2     32 

There  is  no  number  that  will  exactly  divide  both  9  and  32  and,  therefore, 
the  fraction  is  reduced  to  its  lowest  terms. 

12.  Reduction  of  Improper  Fractions. — When  the  numerator 
of  a  fraction  is  just  equal  to  the  denominator,  we  know  that  the 
value  of  the  fraction  is  1  (see  Art.  6) : 

8=1  64  =  1  !?  =  i 

8  64  10 

In  each  of  these  cases  we  have  taken  the  full  number  of  parts 
into  which  we  have  divided  the  unit.  Consequently,  each  of 
these  fractions  is  one  whole  unit,  or  1. 

When  the  numerator  is  greater  than  the  denominator,  the 
value  of  the  fraction  is  one  or  more  units,  plus  a  proper  fraction, 
or  a  whole  plus  some  part  of  a  whole. 

Examples : 

12     8.4     .4        ,1 

-8-  =  8  +  8  =  18°rl2 

47  =  36     11       11 

12     12  +  12       12 

From  these  examples  we  may  see  that  to  reduce  an  improper 
fraction  to  a  whole  or  mixed  number  the  simplest  way  is  as 
follows: 

Divide  the  numerator  by  the  denominator.  The  quotient  will 
be  the  number  of  whole  units.  If  there  is  anything  left  over,  or 
a  remainder,  write  this  remainder  over  the  denominator  since 
it  represents  the  number  of  parts  left  in  addition  to  the  whole 
units.  We  now  have  a  mixed  number,  or  an  exact  whole  number, 
in  place  of  the  improper  fraction. 

Examples : 

27_27.7_Q6  7)27(3 

y=27^7=37  21 

6 

45  3       1  6)45(7 

-6=45^6  =  76       2  f2 

3 


6  SHOP  ARITHMETIC 

These  show  that  a  fraction  represents  unperformed  division. 
In  fact,  division  is  often  indicated  in  the  form  of  a  fraction.  The 
numerator  is  the  dividend  and  the  denominator  is  the  divisor. 

24 

24 -f- 3  can  be  written  — 
o 

2 
2-7-8  can  be  written  = 


13.  Reduction  of  Mixed  Numbers.  —  It  is  often  necessary  or 
desirable  to  change  mixed  numbers  to  improper  fractions.  The 
method  of  doing  this  may  be  seen  from  the  following  examples. 

Examples  : 

Reduce  5-  to  an  improper  fraction. 

4-+J 

In  one  unit  there  are  two  halves.     Therefore, 

,     5X2     10 


1  =  10     1=11 

2  22      2 

If  7  ^  were  to  be  reduced  to  an  improper  fraction  we  would  say:     "Since 

there  are  4  fourths  in  1,  in  7  there  are  4X7,  or  28  fourths.     28  fourths 
plus  1  fourth  equals  29  fourths." 

1=28     1=29 
4~  4      4~  4 

The  rule  which  this  gives  us  is  very  simple:  Multiply  the  whole 
number  by  the  denominator  of  the  fraction  and  write  the  prod- 
uct over  the  denominator.  This  reduces  the  whole  number  to 
a  fraction.  Add  to  this  the  fractional  part  of  the  mixed  number. 
The  sum  is  the  desired  improper  fraction. 

In  working  problems  like  the  above,  the  work  should  be 
arranged  as  in  the  following  example. 


Example : 


Reduce  55  in.  to  eighths  of  an  inch, 
o 


_  1     40     1     41 

5  H  =  ~Q~  +  fi  =  ^~'  Answer- 
o       o       o       o 


COMMON  FRACTIONS  7 

USEFUL  TABLES 

Measures  of  Length 
12    inches  (in.)         =  1  foot  (ft.) 
3    ft.  or  36  in.        =  1  yard  (yd.) 
5J  yd.  or  164  ft.     =  1  rod  (rd.) 
320    rd.  or  5280  ft.    =  1  mile  (mi.) 

Measures  of  Time 

60    seconds  (sec.)  =1  minute  (min.) 
60  minutes  =1  hour  (hr.) 

24    hours  =1  day  (da.) 

7    days  =1  week  (wk.) 

365J  days  =;  1  average  year  (yr.) 

100    years  =  1  century 

Note. — Thirty  days  are  generally  considered  as  one  month, 
though  the  number  of  days  differs  for  different  months. 

Miscellaneous  Units 

12  things  =1 -dozen  (doz.) 

12  dozen  or  144  things  =  1  gross  (gr.) 

12  gross  = 1  great  gross 

20  things  =  1  score 

QUESTIONS  AND  PROBLEMS 

1.  What  is  a  fraction? 

2.  Name  some  fractions  of  an  inch  commonly  used. 

3.  Write  the  following  as  fractions  or  mixed  numbers. 

Five-sixteenths 
Nine  thirty-seconds 
Twenty  and  one-eighth 
Twenty-one  eighths 
Three  and  three-fourths 

4.  Write  out  in  words  the  following: 

ol      5       7       8      5  1     21 

36'    8'     16'     8'    24'    2°~4'     4 

6.  Indicate  the  proper  fractions,  the  improper  fractions,  and  the  mixed 
numbers  among  the  following: 

3    _1      16     21       5       7       9 
4'  6~&'    16'     16'  ^e'    8'     16 

6.  Change  yg  of  an  inch  to  eighths  of  an  inch. 

4 

Change       of  an  inch  to  fourths  of  an  inch. 


8  SHOP  ARITHMETIC 

g 

7.  How  many  sixteenths  of  an  inch  in  j  in.? 

3 

How  many  thirty-seconds  of  an  inch  in  -r  in.? 

13  7  33 

8.  Which  is  greater,  -r^  in.  or  n  in.?    77j  or  To  ? 

9.  Reduce  the  following  mixed  numbers  to  improper  fractions: 

,1      ^1      o3      „!      ,1 
V    V    V    V    '16 

10.  Reduce  the  following  improper  fractions  to  whole  or  mixed  numbers: 

21      8      24     7      121 
16'    8*     3'    2'      12 

11.  I  want  to  mix  up  a  pound  of  solder  to  be  made  of  5  parts  zinc,  2 
parts  tin,  and  1  part  lead.     What  fraction  of  a  pound  of  each  metal — zinc, 
tin  and  lead — must  I  have? 

12.  If  a  train  is  running  at  the  rate  of  a  mile  a  minute,  how  many  feet 
does  it  go  in  1  second? 

7 
•    13.  An  apprentice  who  is  drilling  and  tapping  a  cylinder  for  ^  in.  studs, 

•   3  1 

tries  a  -7  in.  drill  but  the  tap  binds,  so  he  decides  to  use  a  drill  TTT  in.  larger. 

W7hat  size  drill  does  he  ask  for? 

14.  The  tubes  in  a  certain  boiler  are  15  ft.  11  in.  long.     How  many  inches 
long  are  they? 

15.  How  many  seconds  in  an  hour?     40  seconds  is  what  fraction  of  an 
hour? 

16.  An  8-ft.  bar  of  steel  is  cut  up  into  16  in.  lengths.     AVhat  fraction  of 
the  whole  bar  is  one  of  the  pieces? 

17.  When  a  man  runs  100  yd.  in  10  seconds,  how  many  feet  does  he  go 
in  1  second? 

18.  Wood  screws  are  generally  put  up  in  boxes  containing  one  gross. 
If  36  screws  are  taken  from  a  full  box  for  use  on  a  certain  job,  what  fraction 
of  the  gross  is  used  on  this  job  and  what  fraction  is  left  in  the  box?     Reduce 
both  fractions  to  their  lowest  terms. 

19.  A  steel  plate  2  ft.  6  in.  wide  is  to  be  sheared  into  four  strips  of  equal 
width.     How  wide  will  each  strip  be  in  inches? 

20.  In  one  plant  all  drawings  are  dimensioned  in  inches,  while  in  another 
all  dimensions  above  2  ft.  are  given  in  feet  and  inches.     If  a  dimension  is 
given  as  89  in.  in  the  first  plant,  how  would  the  same  dimension  be  stated 
in  the  other  plant? 


CHAPTER  II 
ADDITION  AND  SUBTRACTION  OF  FRACTIONS 

14.  Common  Denominators.  —  Fractions  cannot  be  added 
unless  they  contain  the  same  kind  of  parts,  or,  in  other  words, 
have  the  same  denominator.  When  fractions  having  different 
denominators  are  to  be  added,  they  must  first  be  reduced  to  frac- 
tions having  a  common  denominator.  A  number  of  fractions  are 
said  to  have  a  common  denominator  when  they  all  have  the 
same  number  for  their  denominators,  f  and  £  cannot  be  added 
as  they  stand,  any  more  than  can  3  bolts  and  5  washers.  Both 
the  fractions  must  be  of  the  same  kind,  that  is,  must  have  the 
same  denominator,  £  may  be  changed  to  f  .  By  making  this 
change,  the  fractions  are  given  a  common  denominator  and  can 
now  be  added.  6  eighths  plus  5  eighths  equals  1  1  eighths,  in  just 
the  same  manner  as  6  inches  plus  5  inches  equals  11  inches. 
The  work  of  this  example  would  be  written  as  follows: 


4X2~8 

6     5     11     ,3 

+    =  ~~=1>  Answer- 


8  is  called  the  Least  Common  Denominator  (L.  C.  D.)  of  f 
and  f  ,  because  it  is  the  smallest  number  that  can  be  used  as  a 
common  denominator  for  these  two  fractions.  In  this  case,  the 
least  common  denominator  is  apparent  at  a  glance;  in  many 
other  cases  it  is  more  difficult  to  find,  especially  if  there  are  several 
fractions  to  be  added.  In  the  case  just  given,  the  denominator 
of  one  fraction  can  be  used  for  the  common  denominator.  When 
we  have  two  denominators  like  5  and  8,  neither  of  them  is  an 
exact  multiple  of  the  other  number,  and  so  neither  can  be  the 
common  denominator.  In  such  a  case,  the  product  of  the  two 
numbers  can  be  used  as  a  common  denominator. 

9 


10  SHOP  ARITHMETIC 

Example  : 


5X8  =  40,  theL.  C.  D. 

3^3X8^24 

5     5X8     40 

5^5X5^25 

8~8X5~40 

24     25     49       9 


We  can  always  be  sure  that  the  product  of  the  denominators 
will  be  a  common  denominator,  to  which  all  the  fractions  can  be 
reduced,  but  it  will  not  always  be  the  least  common  denominator. 
For  example,  if  we  wish  to  add  T5^-  and  -j^-,  we  can  use  12X16  = 
192  for  the  common  denominator,  but  we  readily  see  that  48 
will  serve  just  as  well  and  not  make  the  fractions  so  cumbersome. 
In  this  case  48  is  the  least  common  denominator. 

15.  To  Find  the  L.  C.  D.  —  If  the  L.  C.  D.  cannot  be  easily  seen 
by  examining  the  denominators,  it  may  be  found  as  follows: 
Suppose  we  are  to  find  the  L.  C.  D.  of  \,  f  ,    f  ,  and  T\-    First 
place  the  denominators  in  a  row,  separating  them  by  commas. 

2)4,  3,  9,  16 
2)2,  3,  9,  8 
3)1,3,9,  4 
1,  1,  3,  4 
L.  C.  D.  =2X2X3X3X4  =  144 

Select  the  smallest  number  (other  than  1)  that  will  exactly 
divide  two  or  more  of  the  denominators.  In  this  case,  2  will 
exactly  divide  4  and  16.  Divide  it  into  all  the  numbers  that  are 
exactly  divisible  by  it,  that  is,  may  be  divided  by  it  without 
leaving  a  remainder.  When  writing  the  quotients  below,  also 
bring  down  any  numbers  which  are  not  divisible  by  the  divisor 
and  write  them  with  the  quotients.  Now  proceed  as  before, 
again  using  the  smallest  number  that  will  divide  two  or  more 
of  the  numbers  just  obtained.  Continue  this  process  until 
no  number  (except  1)  will  exactly  divide  more  than  one  of  the 
remaining  numbers.  The  product  of  all  the  divisors  and  all  the 
numbers  (except  1's)  left  in  the  last  line  of  quotients  is  the  Least 
Common  Denominator. 

16.  To  Reduce  to  the  L.  C.  D.  —  Having  found  the  least  common 
denominator  of  two  or  more  fractions,  the  next  step  is  to  reduce 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS  11 

the  given  fractions  to  fractions  having  this  least  common 
denominator.  Let  us  take  $,  f,  ±,  and  £•  We  first  find  the 
L.  C.  D.,  which  turns  out  to  be  120.  We  next  proceed  to  reduce 
the  fractions  to  fractions  having  the  L.  C.  D.  Divide  the  com- 
mon denominator  by  the  denominator  of  the  first  fraction. 
Multiply  both  numerator  and  denominator  of  the  fraction  by 
the  quotient  thus  obtained.  Do  this  for  each  fraction,  as  illus- 
trated here. 

2)3,  5,  4,  8 

2)3,  5,  2,  4 

3,  5,  1,  2 

L.C.D.  =2X2X3X5X2  =  120 
1      1X40      40 


3     3X40      120 


120-i-3  = 


120.5  =  24     rsx24-120 
120.4  =  30    i=1X3°      30 


120.8  = 


4  4X30  120 
3  =  3X15^  45^ 
8~~8X15~  120 


17.  Addition  of  Fractions. — Addition  of  fractions  is  very 
simple  after  the  fractions  have  been  reduced  to  fractions  with 
a  common  denominator.  Having  done  this  it  is  only  necessary 
to  add  the  numerators  and  place  this  sum  over  the  common 
denominator.  The  sum  should  always  be  reduced  to  lowest 
terms  and  if  it  turns  out  to  be  an  improper  fraction  it  should 
be  reduced  to  a  mixed  number. 


Example : 


Find  the  sum  of  _5^     3      9       7 
16     4     32     32 
Common  denominator  =32 


. 
16~16X2~32 

3  3X8  =  24 

4  4X8       32 

10  24  _9_  7  =50 
:u  32  32  32  32 
50  25  .9 


If  there  are  mixed  numbers  and  whole  numbers,  add  the  whole 
numbers  and  fractions  separately.     If  the  sum  of  the  fractions 


12  SHOP  ARITHMETIC 

is  an  improper  fraction,  reduce  it  to  a  mixed  number  and  add 
this  to  the  sum  of  the  whole  numbers. 

Example : 

How  long  a  steel  bar  is  needed  from  which  to  shear  one  piece 
each  of  the  following  lengths: 

1731 
7S  in.,  5^  in.,  47  in.,  65  in.  ? 
2  o  4  o 

71      4 
?2     8 

57     7 

Explanation:  The  sum  of  the  whole  numbers  is  22. 
,3      6  18 

47      o  The  sum  of  the  fractions  is  -^-t  which  reduces  to  2-r- 

o  4 

A*      1  Adding  this  to  the  sum  of  the  whole  numbers  (22), 

8      8  1 

gives  24  -r  as  the  sum  of  the  mixed  numbers.    Hence 
22       18     ,-.1  4 

—  ==  2  ~~  1 

2\    84  we  must  have  a  bar  24-  inches  long. 

4 

24-r'  Answer. 
4 

18.  Subtraction  of  Fractions. — Just  as  in  addition,  the  fractions 
must  first  be  reduced  to  a  common  denominator.  Then  we  can 
subtract  the  numerators  and  write  the  result  over  the  common 
denominator. 

Example : 

Subtract    -  from  -=-• 
o  16 

Common  denominator  =  16 
5  =  10 
8     16 

15  10      5 

—  -T^=TS'  Answer. 

16  16     16 

In  subtracting  mixed  numbers,  subtract  the  fractions  first 
and  then  the  whole  numbers. 


Example : 
in.  long? 


1 
How    much    must  be  cut  from  a  15    in.  bolt  to  make  it 


L.  C.  D.  =16 


7^'  Answer. 
Ib 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS  13 

Sometimes,  in  subtracting  mixed  numbers,  we  find  that  the 
fraction  in  the  subtrahend  (the  number  to  be  taken  away)  is 
larger  than  the  fraction  in  the  minuend  (the  number  from  which 
the  subtrahend  is  to  be  taken) .  In  this  case,  we  borrow  1  from 
the  whole  number  of  the  minuend  and  add  it  to  the  fraction  of 
the  minuend.  This  makes  an  improper  fraction  of  the  fraction 
in  the  minuend  and  we  can  now  subtract  the  other  fraction 
from  it. 

Example : 

3  i 

Take  9  7  from  125 

4  8 

19  3 

125  =  115  Explanation:  -7  cannot  be  substracted  from 

o  o  4 

1  /      8\ 

Q3  =     6  -,  so  we  borrow  1  (or^)  from  12  and  write  the 

4      _8  9 

3  minuend   11~- 

25  >  Answer. 
o 


(  or^j 


If  the  minuend  happens  to  be  a  whole  number,  borrow  1  from 
it  and  write  it  as  a  fractional  part  of  the  minuend.  Then  sub- 
tract as  before. 


Example  : 


10-4-T 


-r^>  Answer. 
ID 


PROBLEMS 

53  9  10   24    9 

21.  Reduce  to  the  L.  C.  D.  jg»  j>  and  ^H-  Answer,  ^  oo»  Q-X- 

22.  Reduce  to  the  L.  C.  D.  |»  |.  ^>  and  ~- 

4    o    1U  lo 

,  10   11         ,  15  109     _13 

23.  Add  25,  y^  and  ^-  Answer,  -^  =  2^- 

24>     +    ++"7 


14 


SHOP  ARITHMETIC 


25.  Add  2g>  5^  and  7^- 

15  3 

26.  Find  the  sum  of  8,  3o»  4g>  and  JTT- 

4  7 

27.  Subtract       from      - 


3  1 

29.  Find  the  difference  between  13j  and  202" 

30.  15-ll|=? 

o 

31.  The  weights  of  a  number  of  castings  are:  412^  lb.,  270^  lb.,  1020  lb., 
75^  lb.,  68^  lb.     What  is  their  total  weight? 

32.  Four  studs  are  required:  2-  in.,  lg  in.,  2-r^  in.,  and  loo  in.  long;  how 

3 

long  a  piece  of  steel  will  be  required  from  which  to  cut  them  allowing  -7  in. 

altogether  for  cutting  off  and  finishing  their  ends? 

33.  Monday  morning  an  engineer  bought  48n  gallons  of  cylinder  oil; 

2 
on  Monday,  Tuesday,  and  Wednesday  he  usedj  gallon  per  day;  on  Thurs- 

7  1 

day  he  used  g  gallon;  and  on  Friday  H  gallon.     How  much  oil  had  he  left 

on  Saturday? 


94-- 


_  2i-l"_ 
"8 

Fia.  3. 


34.  Find  the  total  length  of  the  roll  shown  in  the  sketch  in  Fig.  3. 

35.  A  piece  of  work  on  a  lathe  is  1  ft.  in  diameter;  it  is  turned  down  in 

3 
five  cuts;  in  the  first  step  the  tool  takes  off  09  in-  from  the  diameter;  then 


•jg  in. ;  then  oo  m-  >  then  ^o  m->  an<^  the  fifth  time  gj  in.    What  is  the  diam- 
eter of  the  finished  piece? 

36.  How  long  must  a  machine  shop  be  to  accommodate  the  following 
machines  installed  in  a  single  line:  lathe,  8?>  ft.  long;  planer,  14^  ft.  long; 

17  1 

milling  machine,  4g  ft.  long;  engine,  7o  ft.  long;  tool  room,  12g  ft.  long? 

Allow  3j  ft.  between  a  wall  and  a  machine,  and  3^  ft.  between  two  machines. 
The  tool  room  is  to  be  placed  at  the  end  of  the  shop. 

37.  In  doing  a  certain  piece  of  work  one  man  puts  in  lq  hours,  a  second 

man  ?>  hour,  a  third  works  2~-  hours,  and  a  fourth  man  works  lj  hours. 
How  long  would  it  take  one  man  to  do  the  work? 

38.  By  mistake,  the  draftsman  omitted  the  thickness  of  the  flange  on  the 
drawing  of  a  gas  engine  cylinder  in  Fig.  4.     From  the  other  dimensions 
given,  calculate  the  thickness  of  the  flange. 


-J- 


•fr* 


1C 


Fia.  4. 


39.  A  millwright  has  to  rig  up  temporarily  a  6  in.  belt  to  be  367g  in.  long. 

In  looking  over  the  stock  of  old  belting  he  finds  the  following  pieces  of  the 

17  3 

right  width;  one  piece  126^  in.  long,  one  142g  in.  long,  and  one  133g  in.  long. 


16  SHOP  ARITHMETIC 

How  many  inches  must  be  cut  from  one  of  the  pieces  so  that  these  pieces  can 
be  laced  together  to  give  the  right  length? 

40.  The  time  cards  for  a  certain  piece  of  work  show  2  hours  and  15 
minutes  lathe  work,  3  hours  and  10  minutes  milling,  1  hour  and  10  minutes 
planing,  and  1  hour  and  15  minutes  bench  work;  what  is  the  total  number  of 
hours  to  be  charged  to  the  job? 


CHAPTER  III 
MULTIPLICATION  AND  DIVISION  OF  FRACTIONS 

19.  A  Whole  Number  Times  a  Fraction. — In  the  study  of 
multiplication,  we  learn  that  multiplying  is  only  a  short  way  of 
adding.     4x7  is  the  same  as  four  7's  added  together.     Either 
4X7,  or  7+7+7+7  will  give  28.     If  we  apply  this  same  principle 
to  the  multiplying  of  fractions,  we  see  that  4  X  |  is  the  same  as 
four  of  these  fractions  added  together. 

4X7  =  V  +  V  =  28 
A8     8^8^8^8      8 

This  shows  that  multiplying  a  fraction  by  a  whole  number  is 
performed  by  multiplying  the  numerator  by  the  whole  number 
and  placing  the  product  over  the  denominator  of  the  fraction. 

In  other  words,  the  size  of  the  parts  is  not  changed,  but  the 
number  of  parts  is  increased  by  the  multiplication.  After 
multiplying,  the  product  should  be  reduced  to  lowest  terms  and, 
if  an  improper  fraction,  should  be  reduced  to  a  whole  or  mixed 
number. 

Example : 

What   would   be   the   total   weight   of    12   brass   castings   each 
weighing  f  of  a  pound? 

f\         *3f\ 

12X~=-r=9  lb..  Answer. 
4      4 

20.  "Of"   Means  "Times."— The  word  "of"  is  often  seen  in 
problems  in  fractions,  as  for  instance,  "What  is  £  of  5  in.?" 
In  such  a  case,  we  work  the  problem  by  multiplying,  so  we  say 
that  "  of"  means  " times."     You  can  see  that  this  is  so  by  taking 
a  piece  of  wood  5  in.  long  and  cutting  it  into  four  equal  parts 
and  then  taking  three  of  these  parts.     These  three  parts  will  be 
|  of  5  in.,  and  by  actual  measurement  will  be  3J  in  long,  so  we 
know  that  £  of  5  =  3f .     Now  see  what  £  times  5  is' 

3vr  15  3 
TXo  =  -r=3j 
4  44 

which  is  the  same  value.     Therefore,  we  see  that  the    word 
"of"  in  such  a  case  signifies  multiplication. 
2  17 


18  SHOP  ARITHMETIC 

21.  A  Fraction  Times  a  Fraction.  —  To  multiply  two  or  more 
fractions  together,  multiply  the  numerators  together  for   the 
numerator  of  the  product  and  multiply  the  denominators  together 
for  the  denominator  of  the  product. 

Example  : 

Multiply  gxf- 

o      o 

7  2_14_  7          Explanation:    The    numerator  of   the  product  is 

8  3~24~T2  obtained  from  multiplying  the  numerators  to- 

gether:   7X2  =  14.     The    denominator    of     the 
product,  in  the  same  manner,  is  8  X  3  =  24.     This 

14  7 

gives  the  product,,  -  »  which  can  be  reduced  to  j-_- 
24  12 

Let  us  see  what  multiplication  of  fractions  really  means, 
and  why  the  work  is  done  as  just  shown.  Suppose  we  are  to 
find  |  of  |  in.  This  means  that  £  of  an  inch  is  to  be  divided  into 
4  equal  parts  and  3  of  these  parts  are  wanted.  If  we  divide 
|  in.  into  4  equal  parts,  each  part  will  be  one-fourth  as  large  as 
|  in.  and,  therefore,  can  be  considered  as  being  made  up  of 
7  parts,  each  one-fourth  as  large  as  |  in.  Then  \  of  f  =-jV- 
Three  of  these  parts  will  naturally  contain  three  times  as  many 
thirty-seconds,  or  -8^-  =  f^.  Therefore: 

?       7    3     7     3X7     21 

4       8~4X8~4X8~32' 

22.  Multiplying  Mixed  Numbers.  —  This  is   one   of  the  most 
difficult  operations  in  the  study  of  fractions,  unless  one  adopts 
a  fixed  rule  and  follows  it  in  all  cases.     The  student  will  have  no 
trouble  if  he  will  first  reduce  the  mixed  numbers  to  improper 
fractions,  and  then  multiply  these  like  any  other  fractions. 

Example  : 

Find  the  product  of  3^  and  2^- 

1     13  1     5 


113     513X565        1 

~~    ~*'  * 


To  multiply  a  mixed  number  by  a  whole  number,  we  can 
reduce  the  mixed  number  to  an  improper  fraction  and  then 
multiply  it;  or  we  can  multiply  the  fractional  part  and  the 
whole  number  part  separately  by  the  number  and  then  add 
the  products. 


MULTIPLICATION  AND  DIVISION  OF  FRACTIONS  19 

Example : 

3 
What  would  be  the  cost  of  ten  J  in.  by  6  in.  machine  bolts  at  1Q 

o 

cents  a  piece? 
First  method: 

10x1^  =  10  X-Q-  =-8-=  13x  =  13  r  cents. 

Second  Method: 

10  3 

Explanation:  First  multiply  10  by  5-    This  gives 

8  30  6  3 

.  or  36.  or  3.-    Set  this  down.     Then  multiply 
6  o  o  4 

i  3 

4  lObyl.     This  gives  10,  and  we  add  this  to  the  3  » 

l.'V*  cents,  Answer,      giving  a  total  of  13^> 

23.  Cancellation. — Very  often  the  work  of  multiplying  frac- 
tions may  be  lessened  by  cancellation,  as  it  avoids  the  necessity 
of  reducing  the  product  to  lowest  terms.  To  get  an  idea  of 
cancellation  we  must  first  understand  what  a  "factor"  is.  A 
Factor  of  a  number  is  a  number  which  will  exactly  divide  it. 
Thus,  2  is  a  factor  of  8,  3  is  a  factor  of  27,  5  is  a  factor  of  35,  etc. 
When  the  same  number  will  exactly  divide  two  or  more  numbers 
it  is  called  a  common  factor  of  those  numbers.  Thus,  2  is  a 
common  factor  of  8  and  12,  because  it  will  divide  both  8  and  12 
without  leaving  a  remainder.  4  is  also  a  common  factor  of  8 
and  12.  Similarly,  7  is  a  common  factor  of  14  and  21. 

This  idea  of  common  factors  we  have  already  used  in  reducing 
fractions  to  lowest  terms.  Thus,  when  we  have  -fa  we  divide 
both  8  and  12  by  4  and  get 

8  =8-h4  =2 
12    12+4    3 

Cancellation  is  a  process  of  shortening  the  work  of  reduction  by 
removing  or  cancelling  the  equal  factors  from  the  numerator 
and  denominator. 

Example : 

Suppose   we   have   several   fractions   to   multiply   together,   as 


?x2x3x21- 

4*3X14X32 


„,,    .  .  .    3X2X  3  X21      378 

The,r  product  is  ——-- 


20  SHOP  ARITHMETIC 

This  is  not  in  its  lowest  terms  so  we  divide  both  numerator  and  denominator 
by  2,  3,  and  7,  and  get 

378-^-2       189-i-3       63-r-7       9 


5376^2     2688*3  ~  896  -h  7  ~  128 

Now,  if    we  had  struck  out  the  common  factors  from  the  numerator  and 
denominator  before  multiplying  the  fractions,  we  would  have  shortened  the 
work  and  our  answer  would  have  been  in  its  lowest  terms  without  reducing. 
Thus: 

1     1  3 

9 


2X1X2X32     128 
2     1       2 

Explanation:  First  the  3  in  the  numerator  is  cancelled  with  the  3  in 
the  denominator.  This  merely  divides  the  numerator  and  denominator 
by  3  at  the  outset  instead  of  waiting  until  the  terms  are  all  multiplied 
together;  and,  as  3  -f-3  =  1,  we  cancel  a  3  from  both  numerator  and  denomin- 
ator and  place  1's  in  their  stead.  Next  we  divide  both  terms  by  2.  The 
gives  1  in  place  of  the  2  in  the  numerator  and  2  in  place  of  the  4  in  this 
denominator.  Next  we  see  that  7  is  a  common  factor  of  the  numerator  and 
denominator,  so  we  divide  the  21  and  14  each  by  7  and  place  3  and  2  in 
their  places.  There  are  no  more  common  factors;  so  we  multiply  together 

9 

the  numbers  we  now  have  and  get  -=-==• 

iZo 

Another  Example  : 

1 
2465 

=  120 

" 


Explanation:  First  we  cancel  250  out  of  500  and  250;  and  then  9  out  of  36 
and  63;  then  7  out  of  42  and  7;  then  10  out  of  50  and  20;  and  finally  2  out 
of  2  and  2.  This  removes  all  the  common  factors  and  we  get  120  for  the 
answer. 


PROBLEMS  IN  MULTIPLICATION 

41.  7X^  =  ? 

42.  8Xg  =  ? 

23         5 

43.  Multiply  ^  by  Tg' 

3          5 

44.  Find  the  product  of  j  and  „• 

45.  What  is  %  of  2^? 

o  o 

46.  What  is  |  of  |  of  ^  of  |? 


MULTIPLICATION  AND  DIVISION  OF  FRACTIONS  21 

24.  Division — The  Reverse  of  Multiplication. — Division  is  just 
the  opposite  of  multiplication  and  this  fact  gives  us  the  cue  to  a 
very  simple  method  of  dividing  fractions. 

To  divide  one  fraction  by  another,  invert  the  divisor  and  then 
multiply.  To  invert  means  to  turn  upside  down.  Invert  f 
and  we  get  f ;  invert  £  and  we  get  . 

Example : 

TV     -J       27  U      3 

Divide  09 by-- 

9 
27^3=?J     *  =  9     J 

8 

3  4 

Explanation:  The  divisor  is  -:•     Inverting  this  gives  -•    In  multiplying, 

9          1 

we  make  use  of  cancellation  to  simplify  the  work,  and  we  get  -  or  1«  for 

o          o 

the  result. 


Suppose  we  have  a  fraction  to  divide  by  a  whole  number;  as 
-h2  =  ? 
Therefore, 


14  2  1 

4-  2  =  ?     2  is  the  same  as  n  •    If  we  invert   this   we   get  ? 
lu  1  2i 


7 
14     2     Ml       7 

- 


25.  Compound  Fractions. — Sometimes  we  see  a  fraction  which 
has  a  fraction  for  the  numerator  and  another  fraction  for  the 
denominator.  This  is  called  a  Compound  Fraction.  If  we 
remember  that  a  fraction  indicates  the  division  of  the  numerator 
by  the  denominator,  we  will  see  that  a  compound  fraction  can  be 
simplified  by  performing  this  division. 

Example : 

27 

W 1  i;it    i s  iii  ? 

T 

4 

27     3 

This  means  the  same  as  „  --*-     and,  therefore,  would  be  solved  as  follows: 

• '—      4 


27  9     1 

27     * 

x 

8     1 


32     27     3     27     *     9      ,1 

---x-1    Answer- 


22  SHOP  ARITHMETIC 

PROBLEMS  IN  DIVISION 

47.  Divide  175  by  |- 

15          7 

48.  Divide  yg  by  ITT- 

49.  Divide  s|  by  5g- 

60    27-A_? 
&0<  32^10"' 

7 
61.  Find  the  quotient  of  21  -J-K- 


62.  -  =  ? 


26.  How  to  Analyze  Practical  Problems. — The  chief  trouble 
that  students  have  in  working  practical  problems  is  in  analyzing 
the  problems  to  find  out  just  what  operations  they  should  use  to 
work  them.  Problems  in  multiplication  or  division  of  fractions 
will  fall  in  one  of  the  three  following  cases: 

1.  Given  a  whole;  to  find  a  part  (multiply). 

2.  Given  a  part;  to  find  the  whole  (divide). 

3.  To  find  what  part  one  number  is  of  another  (divide). 

Example  of  Case  I : 

The  total  weight  of  a  shipment  of  steel  bars  is  3425  Ib.     ^  of  this  consists 

&O 

3  1 

of  j  in.  round  bars  and  the  balance  is  ?>  in.  round.     What  weight  is  there 

of  each  size? 

137  Explanation :  In   this   example    we 

7       3>f2S      _„.,      .3.     ,  have  the  whole  (3425  Ib.) ;    to  find  a 

^  X  — ~  =959  Ib.  of  j  in.  bars.  /  7  \ 

*  part  \-~-A  •    If  the  whole  is  3425,  then 

3425  -959  =  2466  Ib.  of  ~  in.  bars.      7        V     ' 

~o{  3425  =  959  Ib.  The  balance,  which 
or  ^o  - 

137  1 

1 8     3423  1  consists  of  «  in.  bars,  will  be  3425  —  959, 

^  X^^  =2466  Ib.  of  5  in.  bars. 

rv          1  *  .,       M1  ,       1        7       ^o       7        lo    ,,, 

or  it  will  be  l-25=25-25  =  25ofthe 

18 
whole,     g  of  3425  =  2466  Ib. 

Example  of  Case  2 : 

3 
The  base  of  a  dynamo  weighs  270  Ib.;  the  base  is  yr  of  the  total  weight; 

find  the  total  weight. 


MULTIPLICATION  AND  DIVISION  OF  FRACTIONS  23 

4-  of  the  whole  =  270  lb.  Explanation:    Here   we   have    a 

part  given  to  find   the  whole,    -jy 
whole  =  270  H-  jj-  of  the  whole  is  270  jb      Thig  means 

90  that   if   the   whole   machine    were 

3      270     11  divided    into   eleven   equal    parts, 

270  -5-jj  =  -y-  X-^  =990  lb.,  Answer.      three  of  these  parts  together  would 

weigh  270  lb.  Then  one  part  would 
weigh  270  H- 3  =90  lb.  Since  there 
are  11  of  these  equal  parts,  the 
whole  machine  weighs 

HX90  =  9901b. 
This  is  the  same  as  dividing  270  by 

g 

the  fraction  vy 

Example  of  Case  3 : 

A  molder  who  is  on  piece  work  sets  up  91  flasks,  but  the  castings  from 
7  of  them  are  defective.  What  fractional  part  of  his  work  does  he  get 
paid  for? 

91 —  7  =  84  sound  castings. '  Explanation:  The  problem  is:  What  part 

84     84-*-?     12  of  91  is  84?     There  are  91  parts  in  his  whole 

=      ~=TT   Answer.  g4 

work  and  he  gets  paid  for  84  parts,  or  Q-T  of 

12 

the  whole.    This  can  be  reduced  to-pr;  which 

is  the  answer. 


PROBLEMS 

2 

53.  A  gallon  is  about  JF  of  a  cubic  foot.     If  a  cubic  foot  of  water  weighs 

62^  lb.,  how  much  does  a  gallon  of  water  weigh? 

2  4 

64.  Aluminum  is  2^  times  as  heavy  as  water;  and  copper  is  8v  times  as 

heavy  as  water.     Copper  is  how  many  times  as  heavy  as  aluminum? 

66.  If  a  certain  sized  steel  bar  weighs  2-=  lb.  to  the  foot,  how  long  must  a 
piece  be  to  weigh  8j  lb.? 

66.  What  is  the  cost  of  a  casting  weighing  387«  lb.  at  4j  cents  a  pound? 

67.  How  many  steel  pins  to  finish  lg  in.  long  can  be  cut  from  an  8  ft. 

3 

rod  if  we  allow  -jg  in.  to  each  pin  for  cutting  off  and  finishing? 

68.  A  certain  piece  for  a  machine  can  be  made  of  steel  or  of  cast  iron. 

3  1 

If  drop  forged  from  steel  it  would  weigh  7j  lb.  and  would  cost  6g  cents 

per  pound.     If  made  of  cast  iron,  it  would  have  to  be  made  much  hetvier 


24  SHOP  ARITHMETIC 

and  would  weigh  14  Ib.  and  cost  2g  cents  per  pound.     Which  would  he  the 
cheaper  and  how  much? 

69.  I  want  to  measure  out  2-r  gallons  of  water,  but  I  have  no  measure  at 

hand.     However,  there  are  some  scales  handy  and  I  proceed  to  weigh  out 

7 
the  proper  amount  in  a  pail  that  weighs  ITT  Ib.     What  should  be  the  total 

weight  of  the  pail  and  the  water,  if  one  gallon  of  water  weighs  8^  Ib.  ? 

60.  I  want  to  cut  300  pieces  of  steel,  each  112  in.  long  for  wagon  tires. 
I  have  in  stock  a  sufficient  number  of  bars  of  the  same  size,  but  they  are 
120  in.  long;  and  I  also  have  a  sufficient  number  235  in.  long.  Which  length 
should  I  use  in  order  to  waste  the  least  material?  Calculate  the  total  num- 
ber of  inches  of  stock  that  would  be  wasted  in  each  case. 


CHAPTER  IV 
MONEY  AND  WAGES 

27.  U.  S.  Money.  —  Nearly  every  country  has  a  money  system 
of  its  own.  The  unit  of  money  in  the  United  States  is  the 
dollar.  To  represent  parts  of  a  dollar,  we  use  the  cent,  which  is 
yfo  of  a  dollar.  Fifty  cents  is  •££$  of  a  dollar;  it  is  also  one-half 
dollar  (y5^  =  J)  .  Likewise,  twenty-five  cents  is  T2^  dollar, 
or  one-quarter  dollar. 

In  writing  United  States  money,  the  dollar  sign  ($)  is  written 
before  the  number;  a  period  called  the  decimal  point,  is  placed 
after  the  number  of  dollars;  following  this  decimal  point  is 
placed  the  number  of  cents. 

Two  dollars  and  seventy  cents  is  written  $     2  .  70 

Fifteen  dollars  and  seven  cents  is  written  $  15.07 

One  Hundred  twenty-five  dollars  is  written  $125.00 

One  dollar  and  twenty-five  cents  is  written  $     1  .  25 

Thirty-five  cents  is  written  $       .  35 

Eight  cents  is  written  $       .  08 


Since  one  cent  is  T^  dollar,  it  follows  that  the  figures  to  the 
right  of  the  decimal  point  represent  a  fraction  of  a  dollar.  These 
figures  are  the  numerator,  and  the  denominator  is  100. 

$  2  .  70  is  the  same  as  $  2-iy0- 
$15.07  is  the  same  as  $15rJ-0- 
$  .  08  is  the  same  as  $ 


MONEY  AND  WAGES  25 

The  first  figure  following  the  decimal  point  can  be  said  to 
indicate  the  number  of  dimes,  because  1  dime  =  10  cents,  and 
this  figure  indicates  the  number  of  tens  of  cents.  Also  this 
number  represents  tenths  of  a  dollar,  because  1  dime  =  10  cents 
=  il<nr  dollar  =11U-  dollar.  The  second  figure  after  the  decimal 
point  indicates  cents,  or  hundredths  of  a  dollar. 

This  decimal  system  of  writing  amounts  of  money  has  great 
advantages  in  performing  the  operations  of  addition,  subtraction, 
multiplication,  and  division,  because  we  can  perform  these  opera- 
tions just  as  if  we  were  dealing  with  whole  numbers,  which  makes 
the  work  much  simpler  than  if  we  had  fractions  to  deal  with. 

28.  Addition. — We  can  add  numbers  made  up  of  dollars  and 
cents  and  carry  forward  just  as  in  simple  addition.  The  number 
of  tens  of  cents  will  represent  dimes  (10  cents  =  1  dime;  30  cents 
=  3  dimes,  etc.)  and  thus  can  be  carried  forward  and  added  into 
the  dime  column.  Likewise,  the  number  of  tens  of  dimes  will 
represent  dollars  (10  dimes  =  1  dollar)  and,  therefore,  this  number 
can  be  added  into  the  dollar  column. 

Example : 

Add  $5.20,  $2.65,  $3.25,  and  $.35. 

Explanation:  Adding  the  cents  column,  we  get 
5  +  5  +  5  +  0  =  15  cents.  Put  down  the  5  and  crary 
the  1  into  the  next  column  (since  15  cents  =  1  dime 
and  5  cents.)  Adding  the  dimes,  we  get  1+3+. 
2  +  6  +  2  =  14  dimes.  Put  down  the  4  and  carry 
the  1  into  the  dollar  column  (since  14  dimes  =  $1.4). 
1+3  +  2  +  5  =  11  dollars,  which  we  put  down  com- 


$11.45,  Answer,  plete.  The  decimal  point  we  now  place  in  the 
sum  exactly  as  it  was  in  the  numbers  added,  so 
that  it  properly  separates  dollars  from  cents. 

The  only  precaution  to  be  observed  is  to  see  that  the  dollars, 
dimes,  and  cents  are  properly  lined  up  vertically  before  adding. 
To  do  this  it  is  only  necessary  to  see  that  the  decimal  points  are 
kept  in  a  straight  vertical  line. 


Example : 

What  is  the  total  cost  of  three  articles  priced  as  follows:  $2.25, 
SI,  and  $1.75? 

Explanation:   Here  the  $1   does  not  have  any 

$2.25  decimal  point  or  cents  after  it,  and  care  should 

be  taken  to  see  that  it  is  put  down  in  the  dollar 

l.~-'>  column  and  not  in  the  cents  column.     $1  can  be 

$5.00,  Answer.       written  $1.00,  if  desired,  to  avoid  any  danger  of  a 

mistake. 


26  SHOP  ARITHMETIC 

29.  Subtraction. — The  same  rules  should  be  followed  in  sub- 
tracting.    If  any  figure  in  the  subtrahend  is  larger  than  the 
corresponding  figure  in  the  minuend,  we  can  borrow  1  from  the 
figure  next  to  the  left,  just  as  in  ordinary  subtraction. 

Example : 

A  man  draws  $24.75  on  pay  day  and  immediately  pays  bills 
amounting  to  $8.86.     How  much  does  he  have  left  to  put  in  the  bank? 

Explanation:    Having   set   down   the   numbers 

properly,  we  subtract  j  ust  as  if  we  were  subtract- 

$24.75  ing  whole  numbers.     The  only  difference  is  that, 

8.86  after  subtracting,   we  put  the  decimal  point  in 

$15  gg   Answer.     tn®    remainder  directly  below  the  other  decimal 

points,   to  separate  the  dollars  and  cents  in  the 

remainder. 

30.  Multiplication. — In  multiplying  an  amount  of  money  by 
any  number,  the  process  is  the  same  as  in  simple  multiplication, 
remembering,  however,  to  keep  the  decimal  point  to  separate  the 
dollars  from  the  cents.     The  reasoning  for  this  is  just  the  same  as 
in  addition.     Multiplying  cents,  gives  cents;  multiplying  dimes 
gives  dimes,  and  multiplying  dollars  gives  dollars.     The  figures 
left  over  are  carried  forward  just  as  in  plain  multiplication, 
because  10  cents  =  1  dime,  and  10  dimes  =  1  dollar. 

Examples : 

1.  What  should  a  machinist  receive  for  finishing  48  gas  engines  pistons 
at  25  cents  each? 

$•25  Explanation:  The  25  cents  is  put  down  as  $.25. 

The   multiplication   of  25  by  48  is  performed  as 
200  usual.     Then  the  decimal  point  is  placed  in  the 

100  product  to  separate  the  dollars  and  cents  by  leaving 

<£12  00  Answer.        the  *wo  places,  counting  from  the  right,  to  repre- 
sent cents. 

2.  If  you  owe  2$  weeks  board  at  $3.25  a  week,  how  much  money  will 
it  take  to  settle  the  bill? 

Explanation:  First,  we  put  down  the  numbers 

$3.25  and  then   multiply  as  if  we  were  simply  multi- 

_2i  plying  325  X2$.     The  product  is  812$,  but  since 


162$  we  are  dealing  with  dollars  and  cents,  we   must 

650  put  the  decimal  point   in  the  product  to  show 

<jg  124  Answer  what  part  of  it  represents  dollars  and  what  part 

cents. 

31.  Division. — In  dividing  an  amount  of  money  by  any  number, 
the  division  is  carried  out  as  in  ordinary  division.  The  decimal 
point  is  then  placed  in  the  quotient  in  the  same  position  (from 
the  right)  that  it  had  in  the  dividend. 


MONEY  AND  WAGES  27 

Example : 

The  weekly  pay  roll  of  a  company  employing  405  men  is  $4880.25. 
What  is  the  average  amount  paid  to  each  man? 

405)$4880.25($12.05,  Answer.  Explanation:   The   division  is   carried 

405  out  just  as  if  we  were  dividing  488025  by 

~830  405.     The    quotient    obtained   is    1205. 

g JO  Then,    since  two   of  the  figures  in  the 

dividend  were  cents,  we  place  the  deci- 
mal  point  in  the  quotient  so  as  to  point 
off  the  cents,  making  the  quotient  read 
$12.05. 

Another  way  of  locating  the  decimal  point  is  to  place  it  in 
the  quotient  as  soon  as  the  number  of  dollars  in  the  dividend  has 
been  divided.  Taking  the  same  example;  we  first  divide  405 
into  488  and  get  1,  with  a  remainder  of  83.  Annexing  the  next 
figure  0  and  dividing  again,  we  get  2  for  the  quotient.  We  have 
now  divided  the  number  of  whole  dollars  (4880)  and  have  $12 
for  the  quotient,  with  a  remainder  of  20.  The  12  is,  therefore, 
the  number  of  whole  dollars  in  the  quotient.  We  now  bring 
down  the  next  figure  (2)  from  the  dividend  and  find  that  405 
will  not  go  into  202,  so  we  have  0  dimes.  Then,  bringing  down 
the  five  cents,  we  get  2025  cents,  which,  divided  by  405,  gives 
just  5  cents.  The  men,  therefore,  get  an  average  of  $12.05 
each,  per  week. 

32.  Reducing  Dollars  to  Cents. — Sometimes  we  find  it  desirable 
to  change  a  number  of  dollars  and  cents  all  into  cents.  To  do 
this,  merely  remove  the  decimal  point  from  between  the  dollars 
and  cents  and  you  will  have  the  number  of  cents.  Every  one 
knows  that: 

$1.00  is  100  cents 
$1.25  is  125  cents 
$  .  25  is  25  cents 

Likewise : 

$  12. 75  is  1275  cents 
$  247 . 86  is  24786  cents 
$1000.00  is  100000  cents 

What  we  have  really  done  in  making  these  changes  is  to 
multiply  the  dollars  by  100  to  get  the  equivalent  cents.  We 
have  taken  a  mixed  number  and  multiplied  it  by  100  because 
there  are  100  cents  in  a  dollar.  This  operation  is  performed 
by  moving  the  decimal  point  two  figures  to  the  right,  or  placing 


28  SHOP  ARITHMETIC 

it  after  the  cents,  where  it  is,  of  course,  useless  and  is  seldom 
written. 

In  many  problems  it  is  quite  desirable  to  change  the  dollars 
to  cents  and  carry  the  work  through  as  cents.  The  following 
example  shows  clearly  such  a  case. 

Example : 

During  one  month  a  foundry  turned  out  312,000  Ib.  of  iron 
castings.  The  total  cost  of  the  iron  used,  including  the  cost  of  melting 
and  pouring  was  $3900.  What  was  the  cost,  in  cents,  of  1  Ib.  of  iron, 
melted  and  poured? 

Explanation:  Since  the  cost 

$3900  =  $3900.00  =  390000  cents.  of  iron,  melted  and  poured,  is 

„ ,. ,.  .78      ,1  but  1  or  2  cents,  we  might  as 

390000-^312000  =  1^2  =  14  cents,  Answer.      well  change  the't?tal  c*st  to 

cents  before  we  divide  by  the 
number  of  pounds.  Then  we 
will  get  the  cost  directly  in 
cents  per  pound,  as  we  want  it. 

33.  Reducing  Cents  to  Dollars. — The  reduction  of  cents  to 
dollars  is  really  performed  by  dividing  the  number  of  cents  by 
100,  since  there  are  100  cents  in  1  dollar. 

217  17 

217cents=    ^dollars  =  $2^  =  $2. 17 

Hence,      2 1 7  cents  =  $2 . 1 7 

This  shows  us  that  the  following  simple  rule  can  be  adopted 
for  this  reduction: 

To  reduce  cents  to  dollars,  place  a  decimal  point  in  the  number 
so  as  to  have  two  figures  to  the  right  of  the  decimal  point. 

34.  The  Mill. — There  is  another  division  of  U.  S.  money  called 
the  mill.     A  Mill  is  one-tenth  of  a  cent  or  one  one-thousandth 
of  a  dollar. 

10  mills  =  1  cent 
100  mills  =  10    cents  =  1  dime 
1000  mills  =  100  cents  =  10  dimes  =  1  dollar 

There  is  no  coin  smaller  than  the  cent  and,  therefore,  the  mill 
is  merely  a  name  applied  in  calculations  where  it  is  desirable 
to  have  some  unit  smaller  than  the  cent.  For  example,  tax 
rates  are  usually  given  in  mills  per  dollar.  A  tax  rate  of  15 
mills  on  the  dollar  would  mean  that  a  person  would  have  to  pay 
15  mills  (or  1£  cents)  on  each  dollar  of  assessed  valuation. 
Cost  accountants  generally  figure  costs  down  as  fine  as  mills 
and  even,  in  some  cases,  to  tenths  of  mills  or  finer. 


MONEY  AND  WAGES  29 

In  sums  of  money  containing  mills,  there  are  three  figures 
following  the  decimal  point.  The  first  and  second  figures  after 
the  decimal  point  indicate  dimes  and  cents,  as  before.  The 
third  figure  indicates  mills. 

$ .  014  is  1  cent  and  4  mills 

It  is  also  14  mills  (since  1  cent  =  10  mills). 

In  multiplying  or  dividing  numbers  containing  mills,  we  must 
place  the  decimal  point  in  the  answer  in  the  same  position, 
that  is,  three  places  from  the  right  of  the  number. 

Example : 

If  your  house  and  lot  were  assessed  at  $2000,  and  the  tax  rate 
was  15  mills  on  the  dollar,  what  would  be  the  amount  of  your  taxes? 
15  mills  =  $.015. 

Explanation:  15  mills  is  1  cent  and  5  mills,  or 

$.015  $.015.     This  is  the  amount  you  must  pay  on  each 

2000  dollar  of  assessed   value.     For  an  assessment  of 

I 30  000  '  Answer  $2000,   you  would  pay  2000  times   $.015,  which 

is  $30. 

35.  Wage  Calculations. — The  chief  use  that  shop  men  have, 
within  the  shop,  for  calculations  concerning  money  is  in  connec- 
tion with  their  time  and  wages.  With  some  wage  systems  the 
calculations  of  one's  earnings  is  comparatively  simple;  with 
others  it  seems  rather  complicated  until  the  systems  are  thor- 
oughly understood.  The  simplest  systems  are,  of  course,  the 
well-known  day-rate  and  piece-rate  systems.  By  the  old  day- 
rate  system,  the  men  are  paid  according  to  the  time  put  in, 
without  any  reference  to  the  work  accomplished.  The  rate 
may  be  so  much  per  hour,  per  day,  or  per  week,  but  the  method 
of  calculating  is  the  same,  and  the  time  keeper  will  use  exactly 
the  same  process  in  each  case.  The  pay  roll  calculations  consist 
merely  in  multiplying  the  number  of  units  of  time  which  each 
man  has  to  his  credit  by  his  rate  per  unit  of  time;  hours  by  rate 
per  hour,  or  days  by  rate  per  day. 

Examples : 

1.  A  machinist  puts  in  106  hours  at  32  J  cents  per  hour.     How 
much  money  is  due  him  from  the  company? 

106 

.  32J  Explanation:  The  amount  due  is  the  product  of 

53  106X32J.    Since  the  amount  will  run  into  dollars, 

212  we  write  the  32$  cents  as  dollars,  $.32  J.    The  prod- 

318  uct  is  $34.45,  which  is  the  amount  due. 

$34.45,  Answer. 


30  SHOP  ARITHMETIC 

2.  The  tool-room  foreman  gets  $6.00  a  day  and  has  worked  12  days. 
What  is  the  amount  due  him? 

12  Explanation:  The  same  process  is  carried  out 

6.00  here  except  that  we  have  the  rate  per  day  times 

$727067  Answer.        the  number  of  days. 

The  piece-work  system,  by  setting  prices  for  certain  pieces  of 
work  and  paying  according  to  the  work  done,  rewards  the  man 
in  exact  proportion  to  the  work  that  he  does.  In  this  system, 
an  account  is  kept  of  the  amount  of  a  man's  work  and  his  pay  is 
calculated  by  multiplying  the  numbers  of  pieces  by  the  piece- 
rates. 

Example : 

The  price  for  assembling  a  certain  sized  commutator  is  55  cents 
each.  If  a  man  assembles  an  order  of  14  of  them,  how  much  does  he 
receive?  What  will  he  average  per  day  if  he  does  the  job  in  three  days? 

550=$   .55 

$   .  55  X 14  =  $7 . 70       Amount  he  receives  for  the  j  ob. 
$7.70-j-  3=  2.56$     Amount  he  averages  per  day. 

There  are  many  other  wage  systems,  too  numerous  to  be  all 
explained  here.  They  all  are  planned  to  take  into  account 
both  the  amount  of  work  a  man  does  and  the  time  that  he  puts  in 
to  do  it.  One  of  these  systems,  the  Premium  System,  is  so  well 
known  and  so  successful  as  to  warrant  brief  mention  here.  In 
its  most  common  form  this  system  is  as  follows: 

The  men  are  all  placed  on  a  time  rate,  usually  a  rate  per  hour. 

A  record  is  kept  of  every  man's  time  and  also  of  the  work  done 
by  him. 

Every  job  has  a  standard  time  (sometimes  called  "  the  limit") 
allowed  for  its  completion. 

On  each  job,  the  man's  actual  time  is  recorded  and  also  the 
limit  for  the  job. 

The  man  is  paid  straight  wages  for  the  time  put  in  on  the  job 
and,  in  addition,  is  paid  a  premium  of  usually  one-half  of  the  time 
that  he  saves  below  the  limit. 

If  it  takes  a  man  a  longer  time  than  the  limit,  he  is  paid  full 
wages  for  the  time  he  puts  in. 

This  system  is  planned  to  satisfy  both  parties  by  giving  the 
efficient  man  an  increased  earning,  and  by  giving  the  firm  a 
share  of  the  time  saved,  thus  giving  them  a  reduced  cost  when- 
ever they  pay  higher  wages. 

Let  us  take  an  example  to  see  how  one's  pay  would  be  figured 
on  this  plan. 


MONEY  AND  WAGES  31 

Example : 

In  one  day,  a  man  whose  rate  is  27£  cents  an  hour,  does  the  fol- 
lowing premium  jobs:  the  first  has  a  limit  of  8  hours  and  is  done  in  5  hours; 
the  second  has  a  limit  of  5  hours  and  is  done  in  3  hours;  the  third  has  a 
limit  of  3  hours  and  is  finished  in  2  hours.  What  is  his  pay  for  the  day? 

5  +  3  +  2  =  10  hours,  actually  put  in. 

8—5=  3  hours  saved  on  the  first  job. 

5—3=  2  hours  saved  on  the  second  job. 

3—2=   1  hour  saved  on  the  third  job. 
3  +2  + 1  =  6  hours  saved  on  the  days  work. 

He  will  get  paid  for  the  10  hours  and,  in  addition,  for  half  of  the  6  hours 

that  he  saved.     Altogether  he  will  be  paid  for  10  +  ^  of  6  =13  hours. 

2i 

13X$.272  =  $3.5?2,  Answer. 

He  gets  $3.58  for  his  10  hours  work  and,  therefore,  makes  a  premium  of 
83  cents.  Meanwhile,  the  company  gets  the  work  done  for  $3.58,  instead 
of  paying  $4.40,  which  it  would  have  cost  if  the  workman  had  taken  the 
full  limit  for  the  work. 

PROBLEMS 

61.  Write  in  figures  the  following  sums  of  money: 

One  dollar  and  twelve  cents 

Two  dollars  and  twenty-five  cents 

Eight  cents 

Fifteen  dollars  and  thirty-seven  and  one-half  cents 

Twenty-fi  ve^mills 

62.  Read  the  following  and  write  them  out  in  words: 

$  2.75  $  .008 

$     .03J  $1.08 

$16.25 

63.  A  young  man  makes  the  following  purchases:  Suit  of  clothes  $25, 
shoes   $3.75,   hat   $2.25,   necktie    50  cents.     What  is  the  total  cost  of  his 
purchases? 

64.  A  certain  job  calls  for  four  f  in.  by  3  in.  machine  bolts,  two  f  in.  by  1 J 
in.  set  screws,  and  two  i  in.  by  2  in.  cap  screws.   What  would  be  the  total 
cost  of  the  bolts  and  screws,  if  the  machine  bolts  are  worth  2J  cents  each, 
the  set  screws  1^  cents  each,  and  the  cap  screws  If  cents  each? 

65.  If  you  bought  a  house  for  $3000  and  it  was  assessed  at  two-thirds 
of  what  it  cost  you,  what  taxes  would  you  have  to  pay  if  the  tax  rate  was 
14  mills  on  the  dollar? 

66.  How  much  would  a  man  who  is  paid  $4.25  a  day  earn  in  a  month  of 
26  working  days? 

67.  If  the  piece  price  for  a  certain  job  is  4  cents,  how  many  pieces  must 
a  man  do  in  one  day  to  make  $5.00? 

68.  An  apprentice  and  a  machinist  are  working  together  on  a  piece- 
work job  and  they  earn  $30.     They  are  prorated  on  the  job,  which  means 
that  the    money  is  divided  according  to  their  day-work   rates.     If  the 
apprentice's  rate  is  10  cents  an  hour  and  the  machinist's  is  30  cents,  what 
fraction  of  the  money  does  each  get,  and  how  much  money  would  each  get? 


32 


SHOP  ARITHMETIC 


69.  A  man  rated  at  25  cents  an  hour  is  working  under  the  premium 
system.     In  one  day  of  9J  hours  he  performs  3  operations  the  limits  of 
which  are  4J  hours  each.     If  he  gets  paid  a  premium  of  half  the  time  saved, 
how  much  will  he  make  for  the  day? 

70.  The  following  figure  represents  a  page  from  a  time  book  of  a  shop 
working  entirely  on  day  work.     Calculate  the  total  time  of  each   man, 
the  amount  of  money  due  him,  and  the  total  pay  roll  for  the  shop. 


SHOP-  TWO  WEEKS 


.  I9I./.... 


NAME 


AL 

RS 


AMOUNT 
FOR  THr 

Z  WEEKS 


7-Lfo 


(O  /O  fcr 


/O 


/o 


to 


10  /O 


.£0 


to 


/o 


to 


fO 


/O 


/o 


fo 


/o 


fO 


fO  10 


to 


'0 


10 


to 


iff.  /&. 


/o 


/o 


10  fO 


/o 


/o 


/O  fO 


/o 


fO 


fO 


fo 


fo 


/o 


10 


to 


10 


fo 


fO 


/o 


to 


/o 


/o 


10 


10 


/o 


/o 


/o 


/c 


/o 


fO 


.35- 


fO 


(0 


to 


/o 


/o 


/o 


/o 


/o 


/o 


/o 


/O  fO 


fo 


.20 


10 


/o 


FIQ.  5. 


/o 


/o 


/o 


/o 


to 


TOTAL. 


CHAPTER  V 
DECIMAL  FRACTIONS 

36.  What  are  Decimals?  —  In  the  old  days,  when  no  machinist 
pretended  to  work  much  closer  than  ^  in.  and  the  micrometer 
was  unknown,  the  mechanic  had  little  use  for  decimals  except  in 
figuring  his  pay.  Now,  however,  we  find  that  micrometer 
measurements  are  used  so  generally  that  a  knowledge  of  decimal 
fractions  is  essential. 

A  Decimal  Fraction  is  merely  a  fraction  having  a  denominator 
of  10,  100,  1000,  or  some  similar  multiple  of  10.  The  denomi- 
nator is  never  written,  however,  but  a  system  similar  to  that  used 
in  writing  U.  S.  money  is  used.  A  decimal  fraction  is  written 
by  first  putting  down  a  period  or  "decimal  point"  and  then 
writing  the  numerator  of  the  fraction  after  the  decimal  point  in 
such  a  manner  that  the  denominator  can  be  understood.  Every- 
thing that  comes  after  the  decimal  point  (to  the  right  of  it)  is  a 
fraction,  or  part  of  a  unit. 

In  writing  sums  of  money,  the  first  figure  after  the  decimal 
point  indicates  dimes  or  tenths  of  a  dollar;  the  second  figure 
indicates  cents,  or  hundredths  of  a  dollar;  the  third  figure,  if 
any,  indicates  mills  or  thousandths  of  a  dollar.  This  system 
has  proved  so  handy  that  it  has  been  extended  to  representing 
fractions  of  any  sort  of  a  unit  (not  necessarily  dollars). 

0 

.08    in.  means  y^  in. 

25   . 

.25    in.  means  T^.  in. 
1UU 

256 
1  .  256  in.  means  Ivrx  in. 


Let  us  take  a  decimal,  say  .253,  and  find  out  its  meaning. 

We  said  that  the  first  figure  was  tenths;  the  second,  hundredths; 

the  third,  thousandths,  and  so  on.     Then  .253  would  be  &  -f 

rib"  +  ToW     This  is  not  a  very  handy  system  unless  there  is 

3  33 


34  SHOP  ARITHMETIC 

some  easier  way  to  read  it.     If  we  reduce  these  to  a  common 
denominator  and  add  them,  we  get: 


,        ____         =-.-|_J!L.        3  253 


10     100     1000     1000     1000     1000     1000 
253 


Then  .  253  is 


1000 


This  shows  that  one  place  to  the  right  of  the  decimal  point  indi- 
cates a  denominator  of  10,  two  places  a  denominator  of  100, 
three  places  a  denominator  of  1000,  and  so  on.  The  number 
at  the  right  of  the  decimal  point  can,  therefore,  be  taken  as 
the  numerator,  and  the  denominator  obtained  as  follows:  Put 
down  a  1  below  the  decimal  point  and  a  cipher  (0)  after  it  for 
each  figure  in  the  numerator.  This  will  give  the  denominator. 
In  the  case  just  given,  we  would  have  i%%l  showing  that  the 
denominator  is  1000. 

In  the  same  manner: 

2 
.  2  is  ^>  or  two-tenths. 

37 
.37  is  y^:>  or  thirty-seven  one-hundredths. 

526 

.  526  is          >  or  five  hundred  twenty-six  one-thousandths. 

2749 
.  2749  is  >  or  two  thousand  seven  hundred 

forty -nine  ten-thousandths. 

42 

.042  is          >  or  forty-two  one-thousandths. 
1000 


The  last  case  (.042)  presents  an  interesting  problem.  Here 
we  have  a  numerator  so  small  in  respect  to  the  denominator 
that  it  is  necessary  to  have  a  cipher,  or  zero  (0)  between  it  and  the 
decimal  point,  in  order  that  the  denominator  can  be  indicated 
correctly.  Let  us  see  how  we  would  go  about  writing  such  a 
common  fraction  into  a  decimal.  Take  -nnnr-  If  we  merely 
wrote  .5  that  would  be  -^  and  would,  therefore,  not  be  right. 
From  the  rule  for  finding  denominators  of  decimals  we  see  that 


DECIMAL  FRACTIONS  35 

there  must  be  as  many  figures  after  the  decimal  point  as  there 
are  ciphers  in  the  denominator.  In  this  case  the  denominator 
(1000)  has  3  ciphers,  so  we  must  have  three  figures  in  our  decimal. 
We,  therefore,  put  two  ciphers  to  the  left  of  the  5  and  then  put 
down  the  decimal  point.  We  now  have  .005,  which  can  be 
easily  seen  to  be  -nmr- 

One  thing  that  must  be  carefully  borne  in  mind  is  that  adding 
ciphers  after  a  decimal  does  not  change  the  value  of  the  fraction. 
.5  is  the  same  in  value  as  .50  or  .500  because  T5T  is  the  same  in 
value  as  T5^  or  -nnnr-  On  tne  other  hand,  ciphers  immediately 
following  the  decimal  point  do  affect  the  value  of  the  fraction, 
as  has  just  been  shown. 

Mixed  numbers  are  especially  easy  to  handle  by  decimals, 
because  the  whole  number  and  the  fraction  can  be  written  out 
in  a  horizontal  line  with  the  decimal  point  between  them.  We 
read  mixed  decimals  just  as  we  would  any  mixed  number; — first 
the  whole  number,  then  the  numerator,  and  lastly,  the 
denominator. 


Example : 

42137.24697 

In  this  example,  42137  is  the  whole  number,  and  .24697  is  the  fraction. 
The  number  reads  "forty-two  thousand  one  hundred  thirty-seven  and 
twenty-four  thousand  six  hundred  ninety-seven  hundred-thousandths. 

The  names  and  places  to  the  right  and  left  of  the  decimal 
point  are  as  follows: 


5  i 

^ 

^        <5 


7654321.      123456 

37.  Addition  and  Subtraction.  —  Knowing  that  all  figures  to 
the  right  of  the  decimal  point  are  decimal  parts  of  1  thing  and 
that  all  figures  to  the  left  are  whole  numbers  and  represent  whole 
things,  it  will  be  seen  readily  that  in  addition  and  subtraction  the 
figures  must  be  so  placed  that  the  decimal  points  come  under 


36  SHOP  ARITHMETIC 

each  other.  As  was  shown  under  U.  S.  Money,  the  operations 
can  then  be  carried  out  just  as  if  we  were  dealing  with  whole 
numbers. 

Examples : 

Addition  Subtraction 

783.5  22.7180 

21.473  1.7042 

804.973  21.0138 

Pay  no  attention  to  the  number  of  figures  in  the  decimal. 
Place  the  decimal  points  in  line  vertically.  You  can,  if  you  desire, 
add  ciphers  to  make  the  number  of  decimal  places  equal  in  the 
two  numbers.  Remember,  however,  that  the  ciphers  must 
be  added  to  the  right  of  the  figures  in  the  decimal.  Proceed  as 
in  ordinary  addition  and  subtraction,  carrying  the  tens  forward 
in  addition  and  borrowing,  where  necessary,  in  subtraction  just 
as  with  whole  numbers. 

38.  Multiplication. — In  multiplication  forget  all  about  the 
decimal  point  until  the  work  is  finished ;  multiply  as  usual  with 
whole  numbers.  Then  point  off  in  the  product  as  many  decimal 
places,  counting  from  the  right,  as  there  are  decimal  places  in  the 
multiplier  and  multiplicand  together. 


Example : 


6 . 685  Multiplicand  (3  places) 
5.2  Multiplier       (1  place) 


13370 
33425  

34.7620    Product        (4  places) 

Since  there  are  three  decimal  places  in  one  number,  and  one  in  the  other, 
we  count  off  in  the  product  four  (3  + 1)  places  from  the  right  and  place  the 
point  between  the  7  and  the  4.  The  last  0  can  be  dropped  after  pointing 
off  the  product,  giving  the  result  34.762  (or  34-,^).  The  reason  for  this 
can  be  seen  from  the  following:  The  whole  numbers  are  6  and  5.  The  result 
must  be  a  little  more  than  6X5  =  30,  and  less  than  7X6  =  42,  since  the 
numbers  are  more  than  6  and  5,  and  less  than  7  and  6.  The  actual  result 
is  34.762. 


The  position  of  the  decimal  point  can  be  reasoned  out  in  this 
way  for  any  example,  but  the  quickest  way  is  to  point  off  from 
the  right  a  number  of  decimal  places  equal  to  the  sum  of  the 
numbers  of  decimal  places  in  the  multiplier  and  multiplicand. 


DECIMAL  FRACTIONS  37 

Examples : 

(a)    .0045X2.7  (b)    .000402x4.26 

.  0045  (4  places)  .  000402  (6  places) 

2.7(1  place)  4 . 26  (2  places) 

:U5  2112 

90  804 

.01215  (5  places)  1608 

.00171252  (8  places) 

In  the  above  examples  it  was  necessary  to  put  ciphers  before 
the  product  in  order  to  get  the  required  number  of  decimal 
places.  To  see  the  reason  for  this  take  a  simple  example  such  as 
.2X-3  The  product  is  .06  or  -j-f-g-,  as  can  be  readily  seen  if 
they  are  multiplied  as  common  fractions  (^X^\  =  ^^i). 
This  checks  with  the  rule  of  adding  the  number  of  decimal  places 
in  the  two  numbers  to  get  the  number  in  their  product.  The 
product  of  two  proper  fractions  is  always  less  than  either  of  the 
fractions,  because  it  is  part  of  a  part. 

39.  Short  Cuts. — If  we  want  to  multiply  or  divide  a  decimal  by 
10,  100,  1000,  or  any  similar  number,  the  process  is  very  simple. 
Suppose  we  had  a  decimal  .145  and  then  moved  the  decimal 
point  one  place  to  the  right  and  made  it  1.45.     The  number 
would  then  be  1-j4^  or  {%%  instead  of  iV4Tsir5  so  we  see  that  mov- 
ing the  decimal  point  one  place  to  the  right  has  multiplied  the 
original  number  by  10.     Therefore,  we  see  that: 

To  multiply  by  10  move  the  decimal  point  one  place  to  the 
right. 

To  multiply  by  100  move  the  decimal  point  two  places  to  the 
rigJiT. 

For  other'  similar  multipliers  move  the  decimal  point  one  place 
to'  the  right  for  each  cipher  in  the  multiplier.  This  process  is 
reversed  in  division,  the  rules  being: 

To  divide  by  10  move  the  decimal  point  one  place  to  the  left. 

To  divide  by  100  move  the  decimal  point  two  places  to  the 
left,  etc. 

Example : 

Reduce  10275  cents  to  dollars. 
10275-^-100  =  $102. 75  (Decimal  point  moved  two  places  to  left). 

40.  Division. — The  division  of  decimals  is  just  as  easy  as  the 
multiplication  of  them  after  one  learns  to  forget  the  decimal 
point  entirely  until  the  operation  of  dividing  is  finished.     Divide 


38  SHOP  ARITHMETIC 

as  in  simple  numbers.  Then  point  off  from  the  right  as  many 
decimal  places  in  the  quotient  as  the  number  of  decimal  places 
in  the  dividend  exceeds  that  in  the  divisor.  In  other  words,  we 
subtract  the  number  of  decimal  places  in  the  divisor  from  the 
number  in  the  dividend  and  point  this  number  off  from  the  right 
in  the  quotient. 

Example :  Explanation:  The  number  of 

Divide    105.587  bv  .93  places  in  the  dividend  is  3,  and 

ooMnc  KC7/-i"iQ  K_L     A „ c  in  the  divisor  2.     Hence,  when 

.93)  105. 587(113.5  +  ,  Answer.          thfi    division   ^  complet'ed)  we 

_  point  off  3-2  =  1  place  in  the 

quotient.     The  +  sign  after  the 

Dividend,  3  places  quotient  means  that  the  division 

328         Divisor,  2  places  ^  not  come  out  even>  but  that 

279       Quotient,  1  place  there  was  a  remainder  and  that 

497  the  quotient  given  is  not  com- 

465  plete.     If  desired,  ciphers  could 

~32       Remainder.  °,e  PJ.ac^  after  the  dividend  and 

the  division  carried  farther,  giv- 
ing more  decimal  places  in  the 
quotient. 

It  makes  no  difference  if  the  divisor  is  larger  than  the  dividend, 
as  in  the  following  example.  In  such  a  case  the  quotient  will 
be  entirely  a  decimal. 

Example : 

22.762-5-84.25  =  ? 

84.25)22.762000(.2701+  or  .2702,  Answer. 
16  850 

5  9120 
5  8975 


14500 
8425 
6075 

Explanation:  The  divisor  being  larger  than  the  dividend,  the  quotient 
turns  out  to  be  an  entire  decimal.  In  this  case  we  will  presume  that  we 
wanted  the  answer  to  four  decimal  places.  We  have,  therefore,  added 
ciphers  to  the  dividend  until  we  have  six  decimal  places.  When  these 
have  all  been  used  in  the  division,  we  have  6  —  2  =  4  places  in  the  quotient. 
The  remainder  is  more  than  half  of  the  divisor,  showing  that  if  we  had 
carried  the  division  to  another  place,  the  next  figure  would  have  been 
more  than  5.  We,  therefore,  raise  the  last  figure  (1)  of  the  quotient  to  2, 
because  this  is  nearer  the  exact  quantity. 

In  stopping  any  division  this  way,  if  the  next  figure  of  the 
quotient  would  be  less  than  5,  let  the  quotient  stand  as  it  is,  but, 
if  the  next  figure  would  be  5  or  more,  as  in  the  example  just 
worked,  raise  the  last  figure  of  the  quotient  to  the  next  higher 
figure. 


DECIMAL  FRACTIONS  39 

Sometimes  the  decimal  places  are  equal  in  dividend  and  divisor, 
as  for  instance,  if  we  divide  .28  by  .07. 

.07).  28 
4 

As  the  numbers  of  decimal  places  in  the  dividend  and  divisor  are 
the  same,  the  difference  between  them  is  zero,  and  there  are  no 
decimal  places  in  the  quotient.  The  answer  is  simply  4.  The 
decimal  point  would  come  after  the  4  where  it  would,  of  course, 
be  useless. 

If  there  are  more  decimal  places  in  the  divisor  than  in  the 
dividend,  add  ciphers  at  the  right  of  the  decimal  part  of  the 
dividend  as  far  as  necessary.  In  counting  the  decimal  places, 
be  sure  to  count  only  the  ciphers  actually  used. 

Examples : 

l.H-.025  =  ?  4.2  +  38.25  =  ? 

.  025)  1 . 000(40,  Answer.  38 . 25)4 . 200000( .  1098  + ,  Answer. 
1  00  3  825 

0  37500 

0  34425 


30750 

30600 

150 

41.  Reducing  Common  Fractions  to  Decimals. — Common 
fractions  are  easily  reduced  to  decimals  by  dividing  the  numer- 
ator by  the  denominator.  In  the  case  of  £,  we  divide  1.0  by  2 
and  get  .5  All  that  is  necessary  is  to  take  the  numerator  and 
place  a  decimal  point  after  it,  adding  as  many  ciphers  to  the 
right  of  the  decimal  point  as  are  likely  to  be  needed,  four  being  a 
common  number  to  add,  as  four  decimal  places  (ten  thousandths) 
are  accurate  enough  for  almost  any  calculations. 

If  -g^-  is  to  be  reduced  to  a  decimal,  the  work  is  simply  an 
example  in  long  division,  the  placing  of  the  point  being  the  main 
thing  to  consider.  Simply  divide  1.00000  by  32.  This  gives 
.03125  or  3125  one  hundred-thousandths. 

32)1.00000(.03125 
96 
40 
32_ 
80 
64_ 

16,0 
160 


40 


SHOP  ARITHMETIC 


42.  Complex  Decimals. — A  complex  decimal  is  a  decimal  with 
a  common  fraction  after  it,  such  as  .12J,  .0312^,  etc.  The  frac- 
tion is  not  counted  in  determining  the  number  of  places  in  the 
decimal.  .12£  is  read  "twelve  and  one-half  hundredths." 
.0312£  is  read  "three  hundred  twelve  and  one-half  ten-thou- 
sandths." To  change  a  complex  decimal  to  a  straight  decimal, 
reduce  the  common  fraction  to  a  decimal  and  write  it  directly 
after  the  other  decimal,  leaving  out  any  decimal  point  between 
them. 


Examples : 


.06^=.  065     .8|=.875     .03g=  .03125 


43.  The  Micrometer. — The  micrometer  is  a  device  to  measure 
to  the  thousandth  of  an  inch  and  is  best  known  to  shop  men  in 
the  form  of  the  micrometer  caliper  shown  in  Fig.  6.  The  whole 
principle  of  the  micrometer,  as  generally  made,  can  be  said  to 
depend  on  the  fact  that  ^V  of  TV~TtW-  The  micrometer,  as 
shown  in  Fig.  6,  is  made  up  of  the  frame  or  yoke  6,  the  anvil  c, 
the  screw  or  spindle  a,  the  barrel  d,  and  the  thimble  e.  The 


0   1   2   3   4   50    1 

- 

20 

e 

15 

a 

— 

inliniiiiHu 

,. 

niiliiii 

— 

) 

r- 

1 

FIQ.  6. 

spindle  a  is  threaded  inside  of  d.  The  thimble  e  is  attached  to 
the  end  of  the  spindle  a.  The  piece  to  be  measured  is  inserted 
between  c  and  a,  and  the  caliper  closed  on  it  by  screwing  a 
against  it.  The  screw  on  a  has  40  threads  to  the  inch,  so  if  it  is 
open  one  turn,  it  is  open  ^  in.,  or  T^,  or  .025.  Along  the 
barrel  d  are  marks  to  indicate  the  number  of  turns  or  the  number 
of  fortieths  inch  that  the  caliper  is  open.  Four  of  these  divisions 
(A)  wnl  represent  one-tenth  of  an  inch,  so  the  tenths  of  an  inch 
are  marked  by  marking  every  fourth  division  on  the  barrel. 


41 

Around  the  thimble  e  are  25  equal  divisions  to  indicate  parts  of  a 
turn.  One  of  these  divisions  on  e  will,  therefore,  indicate  -^  of  a 
turn,  and  the  distance  represented  will  be  -^  of  ^  =  T^ns  m- 

To  read  a  micrometer,  first  set  down  the  number  of  tenths  inch 
as  shown  by  the  last  number  exposed  on  the  barrel.  Count  the 
number  of  small  divisions  on  the  barrel  which  are  exposed  be- 
tween this  point  and  the  edge  of  the  barrel.  Multiply  this 
number  by  .025  and  add  to  the  number  of  tenths.  Then  observe 
how  far  the  thimble  has  been  turned  from  the  zero  point  on  its 
edge.  Write  this  number  as  thousandths  of  an  inch  and  add 
to  the  reading  already  obtained.  The  result  is  the  reading  in 
thousandths  of  an  inch. 

Example : 

Let  us  read  the  micrometer  shown  in  Fig.  6. 

.  7  Explanation:  First  we  find  the  figure  7  exposed  on 

.025  the  barrel,  indicating  that  we  have  over  ^  in.     This 

.018  we  put  down  as  a  decimal.     In  addition,  there  is  one 

7743  in.  Answer.     °f  *ne  smaller  divisions  uncovered.     This  is   .025  in 

more.     And    on    the    thimble,    we   find    that  it  is  3 

divisions  beyond  the  15  mark  toward  the  20  mark. 

This  would  be  18,  and  indicates  .018  in.  more.  Adding  the  three, 
.7 +  .025 +  .018  =  .743  in.,  Answer.  This  can  perhaps  be  better  under- 
stood as  being  7  thousandths  less  than  ^  in.  Lots  of  men  locate  a 
decimal  in  their  minds  by  its  being  just  so  far  from  some  common  fraction. 

Most  micrometers  have  stamped  in  the  frame  the  decimal 
equivalents  of  the  common  fractions  of  an  inch  by  sixty-fourths 
from  -fa  in.  to  1  in.  A  table  of  these  decimal  equivalents  is  given 
in  this  chapter,  and  will  be  found  very  useful.  Everyone 
should  know  by  heart  the  decimal  equivalents  of  the  eighths, 
quarters,  and  one-half,  or,  at  least,  that  one-eighth  is  .125.  Then 
f  =  5  X.  125  =  .625;  and  $  =  7 X.  125  =  .875,  etc.  Also,  if  possible, 
learn  that  -^-  =  .062^,  or  .0625.  To  get  the  decimal  equivalent 
of  a  number  of  sixteenths,  add  .062^  to  the  decimal  equivalent 
of  the  eighths  next  below  the  desired  sixteenths. 

Example : 

13 
What  is  the  decimal  equivalent  of  -^  in.? 

||  =  |+^=. 750+. 062^=. 812 Jin.,  or  .8125  in. 

To  set  a  micrometer  to  a  certain  decimal,  first  unscrew  the 
thimble  until  the  number  is  uncovered  on  the  barrel  correspond- 
ing to  the  number  of  tenths  in  the  decimal.  Divide  the  remainder 
by  .025.  The  quotient  will  be  the  additional  number  of  the 


42 


SHOP  ARITHMETIC 


divisions  to  be  uncovered  on  the  barrel  and  the  remainder  will 
give  the  number  of  divisions  that  the  thimble  should  be  turned 
from  zero. 

Example : 

Calculate  the  setting  for  ft  in.  (=  .4375  or  .437$). 

First  unscrew  the  micrometer  until  the  4.  is  uncovered  on  the  barrel. 
Then  divide  the  remainder  .0375  by  .025.  This  gives  1  and  leaves  a  re- 
mainder of  .0125.  The  thimble  should,  therefore,  be  unscrewed  one  full 
turn  or  1  division  beyond  4  on  the  barrel,  plus  12.5  divisions  on  the  thimble. 

TABLE  OF  DECIMAL  EQUIVALENTS  FROM  &  TO  1. 


Fraction 

Decimal 
Equivalent 

Fraction 

Decimal 
Equivalent 

&.... 

i 

.015625 
.03125 

||.... 
4J.  . 

.515625 
.53125 

, 

.046875 
.0625 

f|.... 
9 

.546875 
.5625 

16      5 

751-  ••• 
A.  • 

.078125 
.09375 

10     II...- 
1|.  . 

.578125 
.  59375 

A---- 

.109375 
.125 

5 

.609375 
.625 

5 
A.  • 

.  140625 
.  15625 

8   ii»v 

SI 

.640625 
.  65625 

||.... 

3 

.171875 
.1875 

«•••• 

11 

.671875 
.6875 

10      i, 

04  •  •  •  ' 
A.  • 

.203125 
.21875 

10   «.... 

.703125 

.71875 

H,... 

1 

.234375 
.25 

if...; 

3 

.734375 
.75 

4         J7 
84  •  •  •  ' 

A. 

.265625 
.28125 

4     I!-. 

64 

.  765625 
78125 

19 
b4  '  '  '  ' 

5 

.296875 
.3125 

«-... 

13 

.  796875 

.8125 

16         S1 
6  4  •  •  ' 

u.  . 

.328125 
.34375 

16     If..-. 
££ 

.828125 
84375 

I!.... 

3 

.359375 
.375 

If---. 

7 

.859375 

875 

8      «-... 

n  

.390625 
.40625 

8    «.... 

n.  . 

.890625 
90625 

7   «;; 

.421875 
.4375 

69 
??••.• 

15 

.921875 
9375 

16         J9 

ft  .  .  .  . 



.453125 
.46875 

16  »:::: 

H 

.953125 
96875 

, 

.484375 
.5 

&.... 

1 

.984375 
1  00000 

2 

DECIMAL  FRACTIONS  43 

PROBLEMS 

71.  Write  the  following  as  decimals: 

One  and  twenty-five  one-hundredths. 

Three  hundred  seventy-five  one-thousandths. 

Three  hundred  and  seventy-five  one-thousandths. 

Sixty-two  and  one-half  one-thousandths. 

Seven  hundred  sixty-five  and  five  one-thousandths. 

72.  Read  the  following  decimals  and  write  them  out  in  words: 

.075 
.137 
100.037 
.121 
1.09375 

73.  Find  the  sum  of  .2143,  783.5,  138.72,  and  10.0041. 

74.  From  241. 70  take  215.875. 

76.  a.  Find  the  product  of  78.8763  X  .462. 
b.  Multiply  21.3  by  .071. 

76.  a.  Divide  187.2421  by  123.42. 
b.  Divide  25  by  .0025. 

77.  Reduce  —  in.  to  a  decimal  and  compare  with  the  table. 

78.  Reduce  or,  in.  to  a  decimal  and  compare  with  the  table. 

79.  Calculate  the  decimal  equivalent  of  HT  in. 

80.  Write  .8125  as  a  common  fraction  and  reduce  it  to  the  lowest  terms. 

81.  If  an  alloy  is  .67  copper  and  .33  zinc,  how  many  pounds  of  each  metal 
x  would  there  be  in  a  casting  weighing  75  lb.? 

v    82.  A  steam  pump  delivers  2.35  gallons  of  water  per  stroke  and  runs  48 
strokes  per  minute;  how  many  gallons  will  it  deliver  in  one  hour? 

83.  The  diameter  of  No.  8  B.  W.  G.  wire  is  .165  in.  and  of  No.  12  wire  is 
.109  in.     What  is  the  difference  in  diameter  of  the  two  wires?     What  do 
•  the  letters  B.  W.  G.  stand  for? 

I  84.  A  machinist  whose  rate  is  27.5  cents  per  hour  puts  in  a  full  day  of 
10  hours  and  also  3  hours  overtime.  If  he  is  paid  "time  and  a  half"  for 
overtime,  how  much  should  he  be  paid  altogether? 

3 

85.  The  depth  of  a  thread  on  a  v  in-  bolt  with  U.  S.  Standard  threads  is 

.065  in.     What  is  the  diameter  at  the  bottom  of  the  threads? 

86.  I  want  5000  ft.  of  $  in.  Q  (square)  steel  bars.     I  find  from  a  table 
that  this  size  weighs  1.914  lb.  per  foot  of  length.     How  many  pounds  must 

I  order  and  what  will  it  cost  at  $1.85  per  100  lb.? 

2  i 

87.  Explain  how  you  would  set  a  micrometer  for  JQQQ  in.  over  g  in. 


\ 


44 


SHOP  ARITHMETIC 


88.  A  28-tooth  7-pitch  gear  has  an  outside  diameter  of  4.286  in.     The 
diameter  at  the  bottom  of  the  teeth  is  3.67  in.     How  deep  are  the  teeth  cut? 

89.  A  2  in.  pipe  has  an  actual  inside  diameter  of  2.067  in.     The  metal  of 
the  pipe  is  .154  in.  thick.     What  is  the  outside  diameter  of  the  pipe? 

90.  Read  the  micrometer  shown  below  in  Fig.  7. 


Fio.  7. 


CHAPTER  VI 
PERCENTAGE 

44.  Explanation. — Percentage  is  merely  another  kind  of 
fractions  or,  rather,  a  particular  kind  of  decimal  fractions,  of 
which  the  denominator  is  always  100.  Instead  of  writing  the 
denominator,  we  use  the  term  "per  cent"  to  indicate  that  the 
denominator  is  100.  When  we  speak  of  "6  per  cent"  we  mean 
T|7  or  .06.  These  all  mean  the  same  thing;  namely,  six  parts  out 
of  one  hundred.  Instead  of  writing  out  the  words  "per  cent" 
we  more  often  use  the  sign  %  after  the  number,  as,  for  instance, 
6%,  which  means  "6  per  cent."  Since  per  cent  means  hun- 
dredths  of  a  thing,  then  the  whole  of  anything  is  100%  of  itself, 
meaning  ]--§-£,  or  the  whole.  If  a  man  is  getting  40  cents  an  hour 
and  gets  an  increase  of  10%,  this  increase  will  be  10%  (or  -j^V  or 
.10)  of  40  cents  and  this  is  easily  seen  to  be  4  cents,  so  his  new 
rate  is  44  cents.  Another  way  of  working  this  would  be  to  say 
that  his  old  rate  is  100%  of  itself  and  his  increase  is  10%  of  the 
old  rate,  so  that  altogether  he  is  to  get  110%  of  the  old  rate. 
Now  110%  is  the  same  as  1.10  and  1.10x40  =  44  cents,  the  new 
rate. 


PERCENTAGE  45 

Any  decimal  fraction  may  be  easily  changed  to  per  cent. 

07      K 

•875-^-87.5%. 

Here  we  first  change  the  decimal  to  a  common  fraction  having 
100  for  a  denominator.  Then  we  drop  this  denominator  and 
use,  instead,  the  per  cent  sign  (%)  written  after  the  numerator. 
This  sign  indicates,  in  this  case,  87.5  parts  out  of  100,  or 

87.5 
100 

The  change  from  a  decimal  to  percentage  can  be  made  without 
changing  to  a  common  fraction  as  was  just  done.  Having  a 
decimal,  move  the  decimal  point  two  places  to  the  right  and 
write  per  cent  after  the  new  number. 

.625=62.5%         .06=6%         1.10=110% 

If  it  is  desired  to  use  a  certain  number  of  per  cent  in  calcula- 
tions, it  is  usually  expressed  as  a  decimal  first  and  then  the 
calculations  are  made.  For  example,  when  figuring  the  interest 
on  $1250  at  the  rate  of  6%,  we  would  first  change  6%  to  .06  and 
multiply  $1250  by  .06  which  gives  $75.00. 

$1250 

.06 

$75.00 

A  common  fraction  is  reduced  to  per  cent  by  first  reducing  it 
to  a  decimal  and  then  changing  the  decimal  to  per  cent. 

Example : 

The  force  in  a  shop  is  cut  down  from  85  men  to  62.     What  per  cent 
of  the  original  number  of  men  are  retained? 

no 

62  is  §|  of  85. 
oo 

'  AO 

^=.729  =  72.9% 
62  is  72. 9%  of  85. 

Therefore,  the  number  of  men  retained  is  72.9%  or  nearly  73%  of  the 
original  number  of  men. 

If  we  want  to  reduce  the  fraction  |  to  per  cent,  we  first  get 
£  =  .125  and  then,  changing  this  decimal  to  per  cent,  we  have 
.125  =  12.5%.  Then  |  of  anything  is  the  same  as  12£%  of  it, 
because 


46 


SHOP  ARITHMETIC 


The  following  table  gives  a  number  of  different  per  cents  with 
the  corresponding  decimals  and  common  fractions: 


Per  cent 

Decimal 

Fraction 

Per  cent 

Decimal 

Fraction 

1% 

.01 

1 

100 

25% 

.25 

25       1 
100      4 

2% 

.02 

2         1 

100  "  50 

33J% 

.88| 

33^      1 
100      3 

2*% 

.025 

2i  ..   1 

100      40 

37}% 

.375 

37J      3 

100  ~  8 

5% 

.05 

5         1 

100     20 

50% 

.50 

50       1 
100      2 

61% 

.0625 

6i  _   1 

100      16 

75% 

.75 

75       3 

100  ~  4 

10% 

.10 

10        1 
100      10 

90% 

.90 

90        9 
100      10 

12J% 

.125 

12^       1 

100  ~    8 

100% 

1.00 

100 

100  ~ 

16f% 

-16| 

163      l 

100       6 

200% 

2.00 

200 
100~  * 

45.  The  Uses  of  Percentage. — In  shop  work,  the  chief  use  of 
percentage  is  to  express  loss  or  gain  in  certain  quantities  or  to 
state  portions  or  quantities  that  are  used  or  unused,  good  or  bad, 
finished  or  unfinished,  etc.  Very  often  we  hear  expressions  like: 
"two  out  of  five  of  those  castings  are  bad;"  or  "nine  out  of  ten 
of  those  cutters  should  be  replaced."  If,  in  the  first  illustration, 
we  wanted  to  talk  on  the  basis  of  a  hundred  castings  instead  of 
five,  we  would  say  "40  per  cent  of  those  castings  are  bad," 
because  "two  out  of  five"  is  the  same  as  f,  =  T\\>  =40%.  And 
in  the  second  case:  "90  per  cent  of  those  cutters  should  be 
replaced."  Here,  "nine  out  of  ten"  =^,=^,=90%.  If  a 
piece  of  work  is  said  to  be  60%  completed,  it  means  that,  if  we 
divide  the  whole  work  on  the  job  into  100  £qual  parts,  we  have 
already  done  60  of  these  parts  or  -££$  of  the  whole. 

If  a  shop  is  running  with  50%  of  its  full  force,  it  means  that 
T'W  or  \  °f  the  full  force  is  working.  If  the  full  force  of  men  is 
1300,  then  the  present  force  is  50%  of  1300  =  .50X1300  =  650. 
If  the  full  force  were  700  men,  then  the  50%  would  be  350. 

Another  very  common  use  of  percentage  is  in  stating  the  por- 
tions or  quantities  of  the  ingredients  going  to  make  up  a  whole. 
We  often  see  formulas  for  brasses,  bronzes,  and  other  alloys  in 


PERCENTAGE  47 

which  the  proportions  of  the  different  metals  used  are  indicated 
by  per  cents.  For  example,  brass  usually  contains  about 
65%  copper  and  35%  zinc.  Then,  in  100  Ib.  of  brass, .there 
would  be  65  Ib.  of  copper  and  35  Ib.  of  zinc.  Suppose,  however, 
that  instead  of  100  Ib.  we  wanted  to  mix  a  smaller  amount,  say 
8  Ib.  The  amount  of  copper  needed  would  be  65  %  or  .65  of  8  Ib. 

.65x8  =  5.20  Ib.,  or  5T\  Ib.,  the  copper  needed. 
.35X8  =  2.80  Ib.,  or  2T8T  Ib.,  the  zinc  needed. 

Sometimes,  in  dealing  with  very  small  per  cents,  we  see  a 
decimal  per  cent  such  as  found  in  the  specifications  for  boiler 
steel,  where  it  is  stated  that  the  sulphur  in  the  steel  shall  not 
exceed  .04%.  Now  this  is  not  4%;  neither  is  it  .04;  but  it  is 
.04%,  meaning  four  one-hundredths  per  cent,  or  four  one- 
hundredths  of  one  one-hundredth.  This  is  -j^-g-  of  T^-  =  y  o  o  o  o  >  so 
if  we  write  this  .04%  as  a  decimal,  it  will  be  .0004.  It  is  a  very 
common  mistake  to  misunderstand  these  decimal  per  cents,  and 
the  student  should  be  very  careful  in  reading  them.  Likewise, 
be  careful  in  changing  a  decimal  into  per  cent  that  the  decimal 
point  is  shifted  two  places  to  the  right. 

46.  Efficiencies. — Another  common  use  of  percentage  is  in 
stating  the  efficiencies  of  engines  or  machinery.     The  efficiency  of 
a  machine  is  that  part  of  the  power  supplied  to  it,  that  the  machine 
delivers  up.     This  is  generally  stated  in  per  cent,  meaning  so 
many  out  of  each  hundred  units.     If  it  requires  100  horse-power 
to  drive  a  dynamo  and  the  dynamo  only  generates  92  horse-power 
of  electricity,  then  the  efficiency  of  the  dynamo  is  i\\  or  92%. 
If  the  engine  driving  a  machine  shop  delivers  250  horse-power 
to  the  lineshaft,  but  the  lineshaft  only  delivers  200  horse-power 
to  the  machines,  then  the  efficiency  of  the  lineshaft  is  ff £  =  .80 
=  80%.     The  other  50  horse-power,   or  20%,  is  lost  in  the 
friction  of  the  shaft  in  its  bearings  and  in  the  slipping  of  the 
belts.     The  efficiencies  of  all  machinery  should  be  kept  as  high 
as  possible  because  the  difference  between   100%  and  the  effi- 
ciency means  money  lost.     The  large  amount  of  power  that  is 
often  lost  in  line  shafting  can  be  readily  appreciated  when  we 
try  to  turn  a  shaft  by  hand  and  try  to  imagine  the  power  that 
would  be  required  to  turn  it  two  or  three  hundred  times  a  minute. 

47.  Discount. — In  selling  bolts,  screws,  rivets,  and  a  great 
many  other  similar  articles,  the  manufacturers  have  a  standard 
list  of  prices  for  the  different  sizes  and  lengths  and  they  give  their 


48  SHOP  ARITHMETIC 

customers  discounts  from  these  list  prices.  These  discounts  or 
reductions  in  price  are  always  given  in  per  cent.  Sometimes 
they  -are  very  complicated,  containing  several  per  cents  to  be 
deducted  one  after  another.  Each  discount,  in  such  a  case,  is 
figured  on  the  basis  of  what  is  left  after  the  preceding  per  cents 
have  been  deducted. 

Example : 

The  list  price  of  i  in.  by  1}  iQ-  stove  bolts  is  $1  per  hundred. 
If  a  firm  gets  a  quotation  of  75,  10  and  10%  discount  from  list  price,  what 
would  they  pay  for  the  bolts  per  hundred? 

100  X  .  75  =  75  cents  Explanation:  75,    10   and   10%  discount  means 

100  —  75  =  25  cents  75%  deducted  from  the  list  price,  then  10%  de- 

1  ducted  from  that  remainder,  then  10%  taken  from 

=  2  ce  the  second  remainder. 

1         1  Starting  with  100  cents,  the  list   price,  we  de- 

25-2^  =  22-  cents        duct  the  first  discount  of  75%.     This  leaves  25 

1  ,  cents.     The  next  discount  of  10%  means  10%  off 

22-  X  .  10  =  2-  cents       from  this  balance.     Deducting  this  leaves  22*  cents. 

Next,  we  take   10%  from  this,  leaving  20\  cents 

22!  _  2!  =  20-  cents       per  hundred  as  the  actual  cost  of  these  stove  bolts. 
24         4 

48.  Classes  of  Problems. — Nearly  all  problems  in  percentage 
can  be  divided  into  three  classes  on  the  same  basis  as  explained 
in  Article  26.  There  are  three  items  in  almost  any  percentage 
problem:  namely,  the  whole,  the  part,  and  the  per  cent.  For 
example,  suppose  we  have  a  question  like  this:  "If  35%  of  the 
belts  in  a  shop  are  worn  out  and  need  replacing,  and  there  are 
220  belts  altogether,  how  many  belts  are  worn  out?"  In  this 
case,  the  whole  is  the  number  of  belts  in  the  shop,  220.  The  part 
is  the  number  of  belts  to  be  replaced,  which  is  the  number  to  be 
calculated.  The  per  cent  is  given  as  35%. 

Any  two  of  these  items  may  be  given  and  we  can  calculate  the 
missing  one.  We  thus  have  the  three  cases: 

1.  Given  the  whole  and  the  per  cent,  to  find  the  part. 

2.  Given  the  part  and  the  per  cent  that  it  is  of  the  whole,  to 
find  the  whole. 

3.  Given  the  whole  and  the  part,  to  find  what  per  cent  the, 
part  is  of  the  whole. 

The  principles  taught  under  common  fractions  will  apply 
equally  well  in  working  problems  under  these  cases,  the  only 
difference  being  that  here  a  per  cent  is  used  instead  of  a  common 
fraction.  In  working  problems,  the  per  cent  should  always  be 
changed  to  a  decimal. 

One  difficulty  in  working  percentage  problems  is  in  deciding 


PERCENTAGE  49 

just  what  number  is  the  whole  (or  the  base,  as  it  is  often  called). 
The  following  illustration  shows  the  importance  of  this. 

If  I  offer  a  man  $2000  for  his  house,  but  he  holds  out  for  $3000, 
then  his  price  is  50%  greater  than  my  offer,  while  my  offer  is 
33^%  less  than  his  price.  The  difference  is  $1000  either  way 
but,  if  we  take  my  offer  as  the  base,  it  would  be  necessary  for  me 
to  raise  it  i,  or  50%,  to  meet  his  price.  On  the  other  hand,  for 
him  to  meet  my  bid,  he  would  only  have  to  cut  his  price  J,  or 


Examples  of  the  three  types  of  problems,  before  mentioned, 
may  help  somewhat  in  getting  an  understanding  of  the  processes 
to  be  used. 

Example  of  Case  i  : 

How  many  pounds  of  nickel  are  there  in  1  ton  of  nickel-steel  containing 
2.85%  nickel?  Explanation:  Here  we  have  the  whole 

1  ton  =  2000  Ib.  (200°  lb-)  and  the  per  cent   (2.85%)  to 

2.  85%  =.0285  find  the  part.    After  changing  the  2.85% 

.0285X2000  =  57  lb.,  Answer.         to  a  decimal  fraction,  the  problem  be- 

comes a  simple    problem  in  multiplica- 
tion of  decimals. 

Example  of  Case  2  : 

The  machines  in  a  small  pattern  shop  require  altogether  12  horse-power 

and  are  to  be  driven  from  a  lineshaft  by  a  single  electric  motor.     If  we 

assume  that  20%  of  the  power  of  the  motor  wfll  be  lost  in  the  line  shaft 

and  belting,  what  size  motor  must  we  install? 

100%  —  20%  =  80%  Explanation:  Here  the  per  cent  given  (20%) 

80%=  .80,  or  .8       is  based  on  the  horse-power  of  the  motor,  which 

12  -T-.  8  =  15,  Answer.  is,  as  yet,  unknown.     The  horse-power  of  the 

motor  is  100%  of  itself  and,  if  20%  is  lost, 
then  the  machines  will  receive  80%,  or  .8  of 
the  power  of  the  motor.  This  is  12  horse- 
power. The  whole  will  be  12  -f-.  8  =  15  horse- 
power. This  is  the  size  of  motor  to  install. 

Example  of  Case  3  : 

If  the  force  in  a  shop  is  increased  from  160  to  200  men,  what  per  cent 
is  the  capacity  of  the  shop  increased? 

200-5-160  =  1.25  Explanation:  The  present  force  is  1}  or 

1.25  =  125%  1.25  of  the  old  force,  because  200  is  «:';;  of 

125%  -100%  =  25%,  Answer.    160,  and  fgg  =  l}.    This  is  the  same  as  125%. 

The  increase  is,  therefore,  25%  of  the  former 
force. 

Note.  —  To  reduce  the  present  force  back  to  the  old  number  would  require 
a  reduction  of  only  20%,  because  now  the  base  is  different  on  which  to  figure 
the  per  cent.  40  +  200=  .20,  or  20%. 

PROBLEMS 

91.  Write  4%  as  a  decimal  and  as  a  common  fraction. 

92.  Write  25%  as  a  common  fraction  and  reduce  the  fraction  to  its  lowest 
terms. 

4 


50  SHOP  ARITHMETIC 

93.  What  per  cent  of  an  inch  is  ^  in.? 

94.  If  there  are  240  men  working  in  a  shop  and  30%  of  them  are  laid  off, 
how  many  men  will  be  laid  off  and  how  many  will  remain  at  work? 

96.  Out  of  one  lot  of  342  brass  castings,  21  were  spoiled  and  out  of  an- 
other lot  of  547,  32  were  spoiled.  Which  lot  had  the  larger  per  cent  of 
spoiled  castings? 

96.  500  Ib.  of  bronze  bearings  are  to  be  made;  the  mixture  is  77%  copper, 
8%  tin,  and  15%  lead.     How  many  pounds  of  copper,  tin,  and  lead  are 
required? 

Note. — This  is  a  standard  bearing  mixture  used  by  the  Pa.  R.  R. 
and  by  some  steam  turbine  manufacturers. 

97.  The  boss  pattern  maker  is  given  a  raise  of  25%  on  Christmas,  after 
which  he  finds  that  he  is  receiving  $130  a  month.     How  much  did  he  get 
per  month  before  Christmas? 

98.  In  testing  a  shop  drive  it  was  found  that  the  machines  driven  by  one 
motor  required  horse-power  as  follows: 

60  in.  mill 3.31  horse-power 

20  in.  lathe 75  horse-power 

48  in.  lathe 2.42  horse-power 

42  in.  by  42  in.  by  12  ft.  planer 4.82  horse-power 

16  in.  shaper 33  horse-power 

The  total  power  delivered  by  the  motor  was  13.65  horse-power.  What 
per  cent  of  the  total  power  was  used  in  belting  and  lineshaft?  What  per 
cent  by  the  machines? 

99.  A  man  who  has  been  drawing  $2.50  a  day  gets  his  pay  cut  10%  on 
May  1,  and  the  following  September  he  is  given  an  increase  of  10%  of  his 
rate  at  this  time.     How  much  will  he  get  per  day  after  September? 

100.  The  following  weights  of  metals  are  melted  to  make  up  a  solder: 
18  Ib.  of  tin,  75  Ib.  of  bismuth,  37.5  Ib.  of  lead,  and  19.5  Ib.  of  cadmium. 
What  per  cent  of  the  total  weight  is  there  of  each  metal? 


CHAPTER  VII 

CIRCUMFERENCES  OF  CIRCLES;  CUTTING  AND  GRINDING 

SPEEDS 

49.  Shop  Uses. — In  the  running  of  almost  any  machine,  judg- 
ment must  be  used  in  order  to  determine  the  speed  which  will 
give  the  best  results.     Lathes,  milling  machines,  boring  mills, 
etc.,  are  provided  with  means  for  changing  the  speed,  according 
to  the  -judgment  of  the  operator.     Emery  wheels  and  grind- 
stones, however,  are  often  set  up  and  run  at  any  speed  which 
the  pulleys  happen  to  give,  regardless  of  the  diameter. 

If  an  emery  wheel  of  large  size  is  put  on  a  spindle  that  has  been 
belted  to  drive  a  smaller  wheel,  the  speed  may  be  too  great  for 
the  larger  wheel  and,  if  the  difference  is  considerable,  the  large 
wheel  may  fly  to  pieces.  Every  mechanic  should  know  how  to 
calculate  the  proper  sizes  of  pulleys  to  use  for  emery  wheels  or 
grindstones,  the  correct  speed  at  which  to  run  the  work  in  his 
lathe,  or  the  most  economical  speeds  to  use  for  belts  and  pulleys. 
A  little  data  on  this  subject  may  be  useful  and  will  afford 
applications  for  arithmetical  principles. 

50.  Circles. — To  understand  what  has  just  been  mentioned,  it 
is  necessary  to  get  a  knowledge  of  circles  and  their  properties. 
The  distance  across  a  circle,  measured  straight  through  the  center, 
is  called  the  Diameter.     Circles  are  generally  designated  by  their 
diameters.     Thus  a  6  in.  circle  means  a  circle  6  in.  in  diameter. 
Sometimes  the  radius  is  used.     The  Radius  is  the  distance  from 
the  center  to  the  edge  or  circumference  and  is,  therefore,  just 
half  the  diameter.     If  a  circle  is  designated  by  the  radius,  we 
should  be  careful  to  say  so.     Thus,  there  would  be  no  misunder- 
standing if  we  said  "  a  circle  of  5-in.  radius" ;  but  unless  the  word 
radius  is  used,  we  always  understand  that  the  measurement 
given  is  the  diameter.     The  Circumference  is  the  name  given  to 
the  distance  around  the   circle,   as  indicated  in  Fig.   8.     The 
circumference  of  any  circle  is  always  3.1416  times  the  diameter. 
In  other  words,  if  we  measure  the  diameter  with  a  string  and  lay 
this  off  around  the  circle,  it  will  go  a  little  over  three  times. 

5  51 


52  SHOP  ARITHMETIC 

This  number  3.1416  is,  without  doubt,  the  most  used  in  practical 
work  of  any  figure  in  mathematics.  In  writing  formulas,  it  is 
quite  common  to  represent  this  decimal  by  the  Greek  letter 
TT  (pronounced  "pi")>  instead  of  writing  out  the  whole  number. 
For  this  reason,  the  number  3.1416  is  given  the  name  "pi." 


FIG.  8. 


Where  it  is  more  convenient  and  extreme  accuracy  is  not 
quired,  the  fractio 
exact  value  3.1416. 


22 
required,  the  fraction  --  may  be  used  for  n  instead  of  the  more 


=  3    =  3.1429 

It,  therefore,  gives  values  of  the  circumference  slightly  too 
large,  but  in  many  cases  it  is  sufficiently  accurate  and  saves  time. 

Examples  : 

1.  What  length  of  steel  sheet  would  be  needed  to  roll  into  a  drum  32  in. 
in  diameter? 

When  rolled  up,  the  length  of  the  sheet  will  become  the  circumference  of  a 
32-in.  circle.  The  circumference  must  be  x  times  32. 

3.1416X32  =  100.5+  in. 

The  length  of  the  sheet  must,  therefore,  be  100£  in.  and,  if  it  is  to  be  lapped 
and  riveted,  we  would  have  to  add  a  suitable  allowance  of  1  in.  or  so  for 
making  the  joint. 

2.  A  circular  steel  tank  measures  37  ft.  8$  in.  in  circumference.     What  is 
its  diameter? 

If  the  circumference  of  a  circle  is  3.1416  times  the  diameter,  then  the 
diameter  can  be  obtained  by  dividing  the  circumference  by  3.1416. 

37  ft.  8^  in.  =  37^  ft.  =  37.7+  ft. 

2i  1 

37.7  -4-  3.1416  =  12  ft.,  Answer. 

61.  Formulas.  —  A  formula,  in  mathematics,  is  a  rule  in  which 
mathematical  signs  and  letters  have  been  used  to  take  the  place 
of  words.  We  say  that  "the  circumference  of  a  circle  equals 


CIRCUMFERENCES  OF  CIRCLES  53 

3.1416  times  the  diameter."     This  is  a  rule.     But  suppose  we 
merely  write 


This  is  the  same  rule  expressed  as  a  formula.  We  have  used  C 
instead  of  the  words  "the  circumference  of  a  circle;"  the  sign  = 
replaces  the  word  "equals;"  the  symbol  it  is  used  instead  of  the 
number  3.1416;  X  stands  for  "times;"  and  D  stands  for  "the 
diameter." 

We  found  in  the  second  example  under  Article  50  that,  when 
the  circumference  is  given,  we  can  obtain  the  diameter  by  divid- 
ing the  circumference  by  n.  As  a  formula  this  would  be  written 

D  =  -     or     D  =  C  +  K 

7T 

This  arrangement  is  useful  when  we  want  to  get  \he  diameters 
of  trees,  chimneys,  tanks,  and  other  large  objects.  We  can 
easily  measure  their  circumferences  and,  by  dividing  by  3.1416, 
we  get  the  diameters. 

Formulas  do  not  save  much,  if  any  space,  because  it  is  necessary 
usually  to  explain  what  the  letters  stand  for.  They  have, 
however,  the  great  advantage  that  intricate  mathematical  opera- 
tions can  be  shown  much  more  clearly  than  if  they  were  written 
out  in  a  long  sentence  or  statement.  One  can  usually  see  in  one 
glance  at  a  formula  just  what  is  to  be  done,  with  the  numbers 
that  are  given  in  the  problem,  to  find  the  quantity  that  is 
unknown. 

Example  : 

What  is  the  circumference  in  feet  of  a  16-in.  emery  wheel? 


C  =  3.1416  X  16  =  50.2656  in. 

50.2656  in.  -T-  12  =  4.1888  ft.,  Answer. 

Explanation:  We  have  the  diameter  given  and  want  to  get  the  circum- 
ference. We,  therefore,  use  the  formula  which  says  that  C  =  nXD.  x  is 
always  3.1416  and  D  in  this  case  is  16  in.  Then  C  comes  out  50.2656  in. 
But  the  problem  calls  for  the  circumference  in  feet.  This  is  T'?  of  the  number 
of  inches,  or  it  is  the  number  of  inches  divided  by  12. 

In  the  work  of  this  chapter,  the  circumferences  of  circles 
are  always  used  in  feet,  and,  consequently,  should  always  be 
calculated  in  feet.  If  we  use  D  in  feet,  we  will  get  C  in  feet,  while, 
if  D  is  in  inches,  C  will  come  out  in  inches.  If  the  diameter  can 
be  reduced  to  exact  feet,  it  is  easiest  to  use  the  diameter  in  feet 
when  multiplying  by  TT,  rather  than  to  reduce  to  feet  after 
multiplying. 


54  SHOP  ARITHMETIC 

Example : 

What  is  the  circumference  of  a  48-in.  fly  wheel? 

48  in.  -5- 12  =  4  ft.,  the  diameter. 

C  =  nXD 

(7  =  3.1416X4  =  12.5664  ft.,  Answer. 

This  is  much  shorter  than  it  would  be  to  multiply  3.1416  by  48  and  then 
divide  the  product  by  12. 

52.  Circumferential  Speeds. — When  a  fly  wheel  or  emery 
wheel  or  any  circular  object  makes  one  complete  revolution, 
each  point  on  the  circumference  travels  once  around  the  circum- 
ference and  returns  to  its  starting-point.  When  the  wheel  turns 
ten  times,  the  point  will  have  travelled  a  distance  of  ten  times 
the  circumference.  In  one  minute,  it  will  travel  a  distance  equal 
to  the  product  of  the  circumference  times  the  number  of  revolu- 
tions per  minute.  The  distance,  in  feet  per  minute,  travelled 
by  a  point  on  the  circumference  of  a  wheel  is  called  its  Circum- 
ferential Speed,  Rim  Speed,  or  Surface  Speed.  It  is  also  some- 
times called  Peripheral  Speed,  because  the  circumference  is 
sometimes  given  the  name  of  periphery.  It  is  the  surface  speed 
by  which  we  determine  how  to  run  our  fly  wheels,  belts,  emery 
wheels,  and  grindstones,  and  what  speeds  to  use  in  cutting 
materials  in  a  machine. 
Written  as  a  formula: 
S  =  CXN 
where : 

S  is  the  surface  speed 

C  is  the  circumference 

N  is  the  number  of  revolutions  per  minute  (R.P.  M.). 

Expressed  in  words  this  formula  states  that  the  surface  speed  of 
any  wheel  is  equal  to  the  circumference  of  the  wheel  multiplied 
by  the  number  of  revolutions  per  minute. 

Example : 

What  would  be  the  rim  speed  of  a  7  ft.  fly  wheel  when  running  at 
210  revolutions  per  minute? 

C  =  xXD  Explanation:  First    we    find    the   circum- 

^_22  ference   of   the  wheel,    by  multiplying    the 

~^f  diameter  by  x.     Here  is  a  case  where  it  is 

S  =  CXN  much  easier  to  use  *f-  for  n  than  to  use  the 

5=^2x210  =  4620  decimal  3.1416,  and  the  result  is  sufficiently 

4620  ft.  per  min.,  Answer.        accurate  for  our  purposes.     We  get  22  ft.  for 

the  circumference.     We  can  now  get  the  rim 
speed,  which  is  equal  to  the  product  of  the 

circumference  times  the  number  of  revolutions  per  minute;  orS  =  CxN. 
C  being  22  ft.  and  N  being  210  revolutions  per  minute,  we  find  that  S  is 
4620  ft.  per  min.  Hence,  the  rim  of  this  fly  wheel  travels  at  a  speed  of  4620 
ft.  per  minute. 


CIRCUMFERENCES  OF  CIRCLES  55 

If  we  have  given  a  certain  speed  which  is  wanted  and  have  the 
circumference  of  the  wheel,  then  the  R.  P.  M.  (revolutions  per 
minute)  will  be  obtained  by  dividing  the  desired  speed  by  the 
circumference.  In  the  example  just  worked,  if  we  want  to  give 
the  fly  wheel  a  rim  speed  of  5280  ft.  per  minute,  it  requires 
no  argument  to  show  that  the  wheel  will  have  to  run  at 
5280-^-22  =  240  revolutions  per  minute.  In  such  a  case,  we 
would  use  our  formula  in  the  form 

A7      S 

N=c 

This  formula  expresses  the  same  relation  as  S  =  NxC,  but  now 
it  is  rearranged  to  enable  us  to  find  the  R.  P.  M.  when  the  rim 
speed  and  the  circumference  are  given. 

Sometimes,  especially  with  emery  wheels,  we  know  the  proper 
surface  speed  and  we  have  an  arbor  belted  to  run  a  certain 
number  of  R.  P.  M.  The  problem  then  is  to  find  the  proper  size 
of  stone  to  order. 

The  desired  speed  divided  by  the  number  of  R.  P.  M.  will  give 
the  circumference,  and  from  this  we  can  figure  the  diameter  of 
the  stone. 

c-s 

C~N 

Here  again  we  have  merely  rearranged  the  formula  S  =  CXN  so 
as  to  be  in  more  suitable  form  for  finding  the  circumference  when 
the  surface  speed  and  the  R.  P.  M.  are  given. 

53.  Grindstones  and  Emery  Wheels. — Makers  of  emery  wheels 
and  grindstones  usually  give  the  proper  speed  for  the  stones  in 
feet  per  minute.  This  refers  to  the  distance  that  a  point  on  the 
circumference  of  the  stone  should  travel  in  1  minute  and  is 
called  the  "surface  speed"  or  the  "grinding  speed." 

The  proper  speed  at  which  to  run  grindstones  depends  on  the 
kind  of  grinding  to  be  done  and  the  strength  of  the  stones. 
For  heavy  grinding  they  can  be  run  quite  fast.  For.  grinding 
edge  tools  they  must  be  run  much  slower  to  get  smooth  sur- 
faces and  to  prevent  heating  the  fine  edges  of  the  tools.  The 
following  surface  speeds  may  be  taken  as  representing  good 
practice: 
Grindstones: 

For  machinists'  tools,  800  to  1000  ft.  per  minute. 
For  carpenters'  tools,  550  to  600  ft.  per  minute. 


56  SHOP  ARITHMETIC 

Grindstones  for  very  rapid  grinding: 

Coarse  Ohio  stones,  2500  ft.  per  minute. 

Fine  Huron  stones,  3000  to  3400  ft.  per  minute. 

Sometimes  the  rule  is  given  for  grindstones  as  follows:  "Run 
at  such  a  speed  that  the  water  just  begins  to  fly."  This 
is  a  speed  of  about  800  ft.  per  minute  and  would  be  a  good 
average  speed  for  sharpening  all  kinds  of  tools. 

Examples : 

1.  A  36-in.  grindstone,   used  for  sharpening  carpenters'   and  pattern- 
makers' tools,  is  run  at  60  R.  P.  M.     Is  this  speed  correct? 

We  must  first  find  the  circumference  and  then  the  surface  speed  to  see  if 
it  falls  between  the  allowable  limits. 

36  in. -i- 12  =  3  ft.,  the  diameter  Explanation:  First  we  find  the 

C  =  7tXD  circumference,  which  comes  out 

C  =  3. 1416X3  =  9.4248  ft.  9.4248ft.  Using  this  and  the  R.  P. 

S  =  CxN  M.,  we  find  S  to  be  565  F.  P.  M. 

5  =  9.4248X60  =  565.488  (feet  per  minute).  As  this  lies  be- 

S  =  565.488  ft.  per  minute.  tween  the  allowed  limits  (550  to  600 

F.  P.  M.)  the  speed  of  the  stone  is 

correct. 

2.  At  what  R.  P.  M.  should  a  50-in.  Huron  stone  be  run  if  it  is  to  be  used 
for  rough  grinding? 

C  —  nXD  Explanation:    First   we   find   the 

C  =  3. 1416X50  =  157.08  in.       circumference  of  the  stone  in  feet, 

C  =  157.08  in.  =  13.09  ft.  which  turns  out  to  be  a  little  over  13 

~~To  ft.     The  proper  speed   is  given  as 

g  3000  to  3400  F.  P.  M.     Trying  3200 

N=Y?  .we  find  that  N  comes  out  246  R. 

TVI™  Q     Qonn  v  P   M  P.  M.     The  stone  should,  therefore, 

Take  S  =  320C  F.  P.  M.  be  be]ted  to  fun  about  24Q  Qr  25Q 

AT  =  3200  =  246.+  R.P.M.       R.P.M. 
lo 


Emery  wheels  are  usually  run  at  a  speed  of  about  5500  ft. 
per  minute.  A  good,  ready  rule,  easy  to  remember,  is  a  speed  of 
a  mile  a  minute.  Most  emery  wheel  arbors  are  fitted  with  two 
pulleys  of  different  diameters.  When  the  wheel  is  new,  the  larger 
pulley  on  the  arbor  should  be  used  and,  when  the  wheel  becomes 
worn  down  sufficiently,  the  belt  should  be  shifted  to  the  smaller 
pulley.  Never  shift  the  belt  on  an  emery  wheel,  however,  with- 
out first  calculating  the  effect  on  the  surface  speed  of  the  wheel. 
Many  serious  accidents  have  been  caused  by  emery  wheels 
bursting  as  a  result  of  being  driven  at  too  great  a  speed.  Before 
cutting  a  new  wheel  on  an  arbor  the  resultant  surface  speed 


CIRCUMFERENCES  OF  CIRCLES  57 

should  be  calculated,  to  see  if  the  R.  P.  M.  is  suitable  for  the  size 
of  the  wheel. 

Example  : 

What  size  wheel  should  be  ordered  to  go  on  a  spindle  running 
1700  R.  P.M.? 

-I 


3-106 


D  =  12  in.  wheel,  Answer. 

Note.  —  A  wheel  of  exactly  12  in.  diameter  would,  at  1700  R.  P.  M.,  have* 
a  surface  speed  of  5340  F.  P.  M.  (1700  XT:  =  5340). 

54.  Cutting  Speeds.  —  Cutting  speeds  on  lathe  and  boring  mill 
work  may  be  calculated  in  the  same  way  that  grinding  speeds  are 
calculated.  The  life  of  a  lathe  tool  depends  on  the  rate  at  which 
it  cuts  the  metal.  This  cutting  speed  is  the  speed  with  which 
the  work  revolves  past  the  tool  and  is,  therefore,  obtained  by 
multiplying  the  circumference  of  the  work  by  the  revolutions 
per  minute.  The  same  formulas  are  used  as  in  the  calculations 
for  emery  wheels  and  grindstones  but,  of  course,  the  allowable 
speeds  are  much  different.  Tables  of  proper  cutting  speeds  are 
given  in  many  handbooks  in  feet  per  minute.  To  find  the 
necessary  R.  P.  M.,  divide  the  cutting  speed  by  the  circumference 
of  the  work. 

The  cutting  speeds  used  in  shops  have  increased  considerably 
with  the  advent  of  the  high  speed  steels.  No  exact  figures  can 
be  given  for  the  best  speeds  at  which  to  cut  different  metals. 
The  proper  speed  depends  on  the  nature  of  the  cut,  whether 
finishing  or  roughing,  on  the  size  of  the  work  and  its  ability  to 
stand  heavy  cuts,  the  rigidity  and  power  of  the  lathe,  the  nature 
of  the  metal  being  cut,  and  the  kind  of  tool  used.  If  the  work 
is  not  very  rigid  it  is,  of  course,  best  to  take  a  light  cut  and  run 
at  rather  high  speed.  On  the  other  hand,  it  is  generally  agreed 
that  more  metal  can  be  removed  in  the  same  time  if  a  moderate 
speed  is  used  and  a  heavy  cut  taken. 

As  nearly  as  any  general  rules  can  be  given,  the  following 
table  gives  about  the  average  cutting  speeds. 


58  SHOP  ARITHMETIC 

CUTTING  SPEEDS  IN  FEET  PER  MINUTE 


Kind 

of  tool 

Material 

Carbon  steel 

High  speed  steel 

Cast  iron                       

30  to    40 

60  to    80 

Steel  or  wrought  iron  

25  to    30 

50  to    60 

Tool  steel.  .  .                         

20  to    25 

40  to    50 

Brass.  .        .  .          .  .          

80  to  100 

160  to  200 

Cutting  speed  per  minute  (in  feet) 


revolutions  per  minute. 


^Circumference  of  work  (in  feet) 

Example  : 

A  casting  is  30  in.  in  diameter.     Find  the  number  of  R.  P.  M. 
necessary  for  a  cutting  speed  of  40  ft.  per  minute. 


94.248     _  _ 

—  ^,^—  =  7.804  ft.,  circumference. 

\.Zi 

N  =  ~  =^  ^7  =  5.09  R.  P.  M.,  Answer. 

0         /  .  oo4 

The  same  principles  apply  to  milling  and  drilling,  except  that 
in  these  cases  the  tool  is  turning  instead  of  the  work.  Conse- 
quently, the  cutting  speeds  are  obtained  from  the  product  of  the 
circumference  of  the  tool  times  its  R.  P.  M. 

In  calculating  the  cutting  speed  of  a  drill,  take  the  speed  of  the 
outer  end  of  the  lip  or,  in  other  words,  the  speed  of  the  drill 
circumference. 

Example  : 

A  ^-in.  drill  is  making  300  revolutions  per  minute;  what  is  the 
cutting  speed? 

3.1416X^  =  1.5708,  circumference  in  inches 

a 


—  =.131  ft.  (nearly) 

.131X300  =  39.3  ft.  per  minute,  cutting  speed. 

55.  Pulleys  and  Belts.  —  If  the  rim  of  a  pulley  is  run  at  too 
great  a  speed,  the  pulley  may  burst.  The  rim  speeds  of  pulleys 
are  calculated  in  the  same  manner  as  are  grinding  and  cutting 
speeds.  A  general  rule  for  cast  iron  pulleys  is  that  they  should 
not  have  a  rim  speed  of  over  a  mile  a  minute  (5280  ft.  per  minute)  . 


RATIO  AND  PROPORTION  59 

This  speed  may  be  exceeded  somewhat  if  care  is  taken  that  the 
pulley  is  well  balanced  and  is  sound  and  of  good  design. 

The  proper  speeds  for  belts  is  taken  up  fully  in  a  later  chapter 
under  the  general  subject  of  belting.  It  is  well,  however,  to 
point  out  now  that  the  speed  at  which  any  belt  is  travelling 
through  the  air  is  practically  the  same  as  that  of  the  rim  of 
either  of  the  pulleys  over  which  the  belt  runs;  and,  if  we  neglect 
the  small  amount  of  slipping  which  usually  occurs  between  a 
belt  and  its  pulleys,  we  can  say  that  the  speed  of  a  belt  is  the 
same  as  the  rim  speed  of  the  pulleys.  It  will  be  seen  from  this 
that  if  two  pulleys  are  connected  by  a  belt,  their  rim  speeds  are 
practically  the  same. 

PROBLEMS 

101.  A  stack  is  measured  with  a  tape  line  and  its  circumference  found  to 
be  88  in.     What  is  the  diameter  of  the  stack? 

102.  An  emery  wheel  16  in.  in  diameter  runs  1300  R.  P.  M.  Find  the  sur- 
face speed. 

103.  The  Bridgeport  Safety  Emery  Wheel  Co.,  Bridgeport,  Conn.,  build 
an  emery  wheel  36  in.  in  diameter  and  recommend  a  speed  of  425 — 450 
revolutions.     Calculate  the  surface  speeds  at  425  and  at  450  revolutions. 

104.  An  emery  wheel  runs  1000  R.  P.  M.     What  should  be  its  diameter 
to  give  a  surface  speed  of  5500  ft.? 

105.  A  grindstone  3 \  ft.  in  diameter  is  to  be  used  for  grinding  carpenters' 
tools;  how  many  R.  P.  M.  should  it  run? 

106.  Calculate  the  belt  speed  on  a  high-speed  automatic  engine  carrying 
a  48  in.  pulley  and  running  at  250  R.  P.  M. 

107.  How  many  revolutions  will  a  locomotive  driving  wheel,  72  in.  in 
diameter,  make  in  going  1  mile? 

108.  What  would  be  the  rim  speed  in  feet  per  minute  of  a  fly  wheel  14 
ft.  in  diameter  running  80  R.  P.  M.? 

109.  At  how  many  R.  P.  M.  should  an  8  in.  shaft  be  driven  in  a  lathe  to 
give  a  cutting  speed  of  60  ft.  per  minute? 

110.  At  what  R.  P.  M.  should  a  1^  in.  high  speed  drill  be  run  to  give  a 
cutting  speed  of  80  ft.  per  minute?     If  the  drill  is  fed  .01  in.  per  revolution, 
how  long  will  it  take  to  drill  through  2  in.  of  metal? 

CHAPTER  VIII 
RATIO  AND  PROPORTION 

66.  Ratios. — In  comparing  the  relative  sizes  of  two  quantities, 
we  refer  to  one  as  being  a  multiple  or  a  fraction  of  the  other.  If 
one  casting  weighs  600  lb.,  and  another  weighs  200  lb.,  we  say 
that  the  first  one  is  three  times  as  heavy  as  the  second,  or  that 


GO  SHOP  ARITHMETIC 

the  second  is  one-third  as  heavy  as  the  first.  This  relation 
between  two  quantities  of  the  same  kind  is  called  a  Ratio. 

In  comparing  the  speeds  of  two  pulleys,  one  of  which  funs  40 
revolutions  per  minute  and  the  other  one  160  revolutions  per 
minute,  we  say  that  their  speeds  are  "  as  40  is  to  160,"  or  "  as  1  is 
to  4."  In  this  sentence,  "40  is  to  160"  is  a  ratio,  and  so  also  is 
"1  is  to  4"  a  ratio. 

Ratios  may  be  written  in  three  ways.  For  example,  the  ratio 
of  (or  relation  between)  the  diameters  of  two  pulleys  which  are 
12  in.  and  16  in.  in  diameter  can  be  written  as  a  fraction,  -J-f  ;  or, 
since  a  fraction  means  division,  it  can  be  written  12-7-16;  or, 
again,  the  line  in  the  division  sign  is  sometimes  left  out  and  it 
becomes  12:16.  The  last  method,  12:16,  is  the  one  most  used 
and  will  be  followed  here.  It  is  read  "twelve  is  to  sixteen." 

A  ratio  may  be  reduced  to  lower  terms  the  same  as  a  fraction, 
without  changing  the  value  of  the  ratio.  If  one  bin  in  the  stock 
room  contains  1000  washers,  while  another  bin  contains  3000, 
then  the  ratio  of  the  contents  of  the  first  bin  to  the  contents  of 
the  second  is  "  as  1000  is  to  3000."  The  relation  of  1000  to  3000 
can  be  reduced  by  dividing  both  by  1000.  This  leaves  the  ratio 
1  to  3. 

1000  -s-  1000     1 


Hence,  the  ratio  between  the  contents  of  the  bins  is  also  as  1  is 
to  3. 

Likewise,  the  ratio  24:60  can  be  reduced  to  2:5  by  dividing 
both  terms  by  12.  If  we  write  it  as  a  fraction  we  can  easily  see 
that 

24  =  24-«-12_2 
60~60-f-12~5 

Therefore,  24:60  =  2:5. 

The  ratio  of  the  1000  washers  to  the  3000  washers  is  1000:3000 
or  1:3. 

The  ratio  of  8  in.  to  12  in.  is  8:12  or  2:3. 

The  ratio  of  $1  to  $1.50  is  1:1  J  or  2:3. 

The  ratio  of  30  castings  to  24  castings  is  30:24  or  5:4. 

57.  Proportion.  —  When  two  ratios  are  equal,  the  four  terms 
are  said  to  be  in  proportion.  The  two  ratios  2:4  and  8:16  are 
clearly  equal,  because  we  can  reduce  8:16  to  2:4  and  we  can 


RATIO  AND  PROPORTION  61 

therefore    write    2:4  =  8:16.     When    written    thus,    these    four 
numbers  form  a  Proportion. 

Likewise,  we  can  say  that  the  numbers  6,  8,  15,  and  20  form 
a  proportion  because  the  ratio  6:8  is  equal  to  the  ratio  of  15:20. 

6:8  =  15:20 

Now,  it  will  be  noticed  that,  if  the  first  and  fourth  terms  of 
this  proportion  be  multiplied  together,  their  product  will  be 
equal  to  the  product  of  the  second  and  third  terms: 

/First\  /SecondN  /Third\  /Fourth\ 

Vterm/  V  term   )  \  term  /  \  term   / 

6:8  15      ,     :          20 

6X20  =  120 
8X15  =  120 
Therefore,  6X 20  =  8X15 

This  is  true  of  any  proportion  and  forms  the  basis  for  an  easy 
way  of  working  practical  examples,  where  we  do  not  know  one 
term  of  the  proportion,  but  know  the  other  three.  The  first  and 
fourth  terms  are  called  the  Extremes,  and  the  second  and  third 
are  called  the  Means.  Then  we  have  the  rule:  "  The  product  of 
the  means  is  equal  to  the  product  of  the  extremes." 

This  relation  can  be  very  nicely  and  simply  expressed  as  a 
formula. 

Let  a,  &,  c,  and  d  represent  the  four  terms  of  any  proportion 
so  that 

a  : b=c  : d 

Then,  according  to  our  rule,  we  have 

aXd=bXc 

Let  us  now  see  of  what  practical  use  this  is.  We  will  take  this 
example: 

If  it  requires  137  Ib.  of  metal  to  make  19  castings,  how  many 
pounds  will  it  take  to  make  13  castings  from  the  same  pattern? 

Now  very  clearly  the  ratio  between  the  number  of  castings 
19:13  is  the  same  as  the  ratio  of  the  weights,  but  one  of  the 
weights  we  do  not  know.  Writing  the  proportion  out  and  put- 
ting the  word  "  answer"  for  the  number  which  we  are  to  find,  we 
have 

19:13  =  137  :Answer 


02  SHOP  ARITHMETIC 

From  our  rule  which  says  the  product  of  the  means  equals  the 
product  of  the  extremes: 

13  X 137  — 1781,  product  of  "  means." 
This  must  equal  the  product  of  the  extremes  which  would  be 

19  X  Answer. 
Then:  19  X  Answer  =  1781 

Answer  =  1781-*- 19 

Answer  =  93. 7+  Ib. 

In  using  proportion  keep  the  following  things  in  mind: 

(1)  Make  the  number  which  is  the  same  kind  of  thing  as  the 
required  answer  the  third  term.     Make  the  answer  the  fourth 
term. 

(2)  See  whether  the  answer  will  be  greater  or  less  than  the 
third  term;  if  less,  place  the  less  of  the  other  two  numbers  for  the 
second  term;  if  greater,  place  the  greater  of  the  other  two  numbers 
for  the  second  term. 

(3)  Solve  by  knowing  that  the  product  of  the  means  equals  the 
product  of  the  extremes,  or  by  this  rule:  Multiply  the  means 
together  and  divide  by  the  given  extreme;  the  result  will  be  the 
other  extreme  or  answer. 

Let  us  see  how  these  rules  would  be  applied  to  a  practical 
example. 

Example : 

A  countershaft  for  a  grinder  is  to  be  driven  at  450  R.  P.  M.  by  a 
lineshaft  that  runs  200  R.  P.  M.  If  the  pulley  on  the  countershaft  is  8  in. 
in  diameter,  what  size  pulley  should  be  put  on  the  lineshaft?  A  proportion 
can  be  formed  of  the  pulley  diameters  and  their  revolutions  per  minute. 
Applying  the  rules  of  proportion,  we  get  the  following  analysis  and  solution 
to  the  problem. 

(1)  The  diameter  of  the  lineshaft  pulley  is  the  unknown  answer.     The 
other  number  of  the  same  kind  is  the  diameter  of  the  countershaft  pulley 
(8  in.).     So  we  have  the  ratio. 

8  :Answer 

(2)  If  the  countershaft  pulley  is  to  run  faster,  its  diameter  must  be 
smaller  than  the  other  one.     Therefore,   the  answer  is  greater  than  8. 
Hence,  the  greater  revolutions  (450)  will  be  placed  as  the  second  term  and 
the  other  R.  P.  M.  (200)  will  be  the  first  term.     Therefore,  we  have  the 
completed  proportion: 

200:450  =  8 :  Answer 

(3)  Solving  this  we  get: 

450X8  =  3600,  product  of  means. 
3600-^-200  =  18,  Answer. 

Hence,  an  18-in.  pulley  should  be  put  on  the  lineshaft  to  give  the  desired 
speed  to  the  countershaft. 


RATIO  AND  PROPORTION  03 

Sometimes  the  letter  X  is  used  to  represent  the  unknown 
number  whose  value  is  sought.  The  following  is  an  example  of 
such  a  case. 

6:40  =  5:X.     Find  what  number  X  stands  for. 
40  X  5  =  200,  product  of  the  means. 

Hence,  6XX  =  200 

200 
X=~ =33.3+,  Answer. 

58.  Speeds  and  Diameters  of  Pulleys. — As  shown  in  an  example 
previously  worked,  if  two  pulleys  are  belted  together,  their 
diameters  and  revolutions  per  minute  can  be  written  in  a  pro- 
portion having  diameters  in  one  ratio  and  R.  P.  M.  in  the  other 
ratio  of  the  proportion.  It  will  be  noticed  from  the  example 
which  was  worked,  that  the  numbers  which  form  the  means  apply 
to  the  same  pulley,  while  the  extremes  both  refer  to  the  other  pul- 
ley. Then,  since  the  product  of  the  means  equals  the  product 
of  the  extremes,  we  obtain  the  following  simple  relation  for 
pulleys  belted  together:  The  product  of  the  diameter  and  revolu- 
tions of  one  pulley  equals  the  product  of  the  diameter  and  revolu- 
tions of  the  other.  This  gives  us  the  following  simple  rule  for 
working  pulley  problems. 

Rule  for  Finding  the  Speeds  or  Diameters  of  Pulleys. — Take  the 
pulley  of  which  we  know  both  the  diameter  and  the  R.  P.  M., 
and  multiply  these  two  numbers  together.  Then  divide  this 
product  by  the  number  that  is  known  of  the  other  pulley.  The 
result  is  the  desired  number. 

Examples : 

1.  A  36-in.  pulley  running  240  R.  P.  M.  is  belted  to  a  15-in.  pulley.     Find 
the  R.  P.  M.  of  the  15-in.  pulley. 

36  X  240  =  8640,  the  product  of  the  known  diameter  and  revolutions. 
8640 -=-15  =  576,  the  R.  P.  M.  of  the  15-in.  pulley,  Answer. 

2.  A  36-in.  grindstone  is  to  be  driven  at  a  speed  of  800  R.  P.  M.  from  a 
6-in.  pulley  on  the  lineshaft  which  is  running  225  R.  P.  M.     \Vhat  size  pulley 
must  be  put  on  the  grindstone  arbor? 

S  Explanation:  First  we  must  find 

C  the  R.  P.  M.  for  the  grindstone  as 

800  explained  in  Chapter  VII.     To  get 

=  Q  .  .--     ~  =85  R.  P.  M.,  nearly,  the  required  surface  speed  we  find 

« v99*     ,o-n  85  R.  P.  M.  necessary. 

,  Now  we  have  the  R.  P.  M.  and 

the   size    of  the  lineshaft  pulley. 
Use  a  16-m.  pulley  on  the  arbor.       The  product  of  these  two  nuJjnber8 

is   1350.     Dividing  this  by  the  R. 

P.  M.  of  the  grindstone  arbor  gives  16  in.  as  the  nearest  even  size  of  pulley, 
so  we  will  use  that  size. 


64 


SHOP  ARITHMETIC 


69.  Gear  Ratios. — The  same  principles  as  are  applied  to  pulleys 
can  be  applied  to  gears.  If  we  have  two  gears  running  together 
as  shown  in  Fig.  9,  the  product  of  the  diameter 'and  R.  P.  M.  of 
one  gear  will  be  equal  to  the  product  of  the  diameter  and  R.  P.  M. 
of  the  other.  In  studying  gearing,  we  do  not  deal  with  the  diam- 
eters so  much  as  we  do  with  the  numbers  of  teeth.  We  find  that 
gears  are  generally  designated  by  the  numbers  of  teeth.  For 
example,  we  talk  of  16  tooth  gears  and  24  tooth  gears,  etc.,  but 
we  seldom  talk  about  gears  of  certain  diameters. 

In  making  these  calculations  for  gears,  we  can  use  the  numbers 
of  teeth  instead  of  the  diameters.  When  a  gear  is  revolving,  the 
number  of  teeth  that  pass  a  certain  point  in  one  minute  will  be 
the  product  of  the  number  of  teeth  times  the  R.  P.  M.  of  the 
gear. 


FIG.  9. 


If  this  gear  is  driving  another  one,  as  in  Fig.  9,  each  tooth  on 
the  one  gear  will  shove  along  one  tooth  on  the  other  one.  Conse- 
quently, the  product  of  the  number  of  teeth  times  R.  P.  M.  of 
the  second  gear  will  be  the  same  as  for  the  first  gear.  This  gives 
us  our  rule  for  the  relation  of  the  speeds  and  numbers  of  teeth  of 
gears. 

Rule  for  Finding  the  Speeds  or  Numbers  of  Teeth  of  Gears.— 
Take  the  gear  of  which  we  know  both  the  R.  P.  M.  and  the  num- 
ber of  teeth  and  multiply  these  two  numbers  together.  Divide 
their  product  by  the  number  that  is  known  about  the  other  gear. 
The  quotient  will  be  the  unknown  number. 


PULLEY  AND  GEAR  TRAINS  65 

Example : 

A  38  tooth  gear  running  360  R.  P.  M.  is  to  drive  another  gear  at 
190  R.  P.  M.     What  must  be  the  number  of  teeth  on  the  other  gear? 

38  X  360  =  13,680,   the  product  of  the  number  of  teeth  and  revolutions 

of  one  gear. 
1 90  X  Answer  =  13, 680 

13,680 
Answer  «- 


190 
Answer  =  72  teeth. 


PROBLEMS 


111.  (a)  If  you  draw  $33.00  on  pay  day  and  another  man  draws  $22.00, 
what  is  the  ratio  of  your  pay  to  hist 

(b)  What  is  the  ratio  of  his  pay  to  yours? 

112.  The  speeds  of  two  pulleys  are  in  the  ratio  of  1:4.     If  the  faster  one 
goes  260  R.  P.  M.,  how  fast  does  the  slower  one  go? 

113.  Two  castings  are  weighed  and  the  ratio  of  their  weights  is  5:2. 
If  the  lighter  one  weighs  80  lb.,  what  does  the  heavier  one  weigh? 

114.  Find  the  unknown  number  in  each  of  the  following  proportions: 

(a)  2:10  =  5:Answer 

(b)  6:42  =  5:Answer 

(c)  7:35  =  10:X 

(d)  6:72  =  8 iX" 

115.  If  it  takes  72  lb.  of  metal  to  make  14  castings,  how  many  pounds  are 
required  to  make  9  castings? 

116.  A  14  tooth  gear  is  driving  a  26  tooth  gear.     If  the  14  tooth  gear 
runs  225  revolutions  per  minute,  what  is  the  speed  of  the  26  tooth  gear? 

117.  A  12  in.  lineshaft  pulley  runs  280  revolutions  and  is  belted  to  a 
machine  running  70  revolutions.     What  must  be  the  size  of  the  pulley  on 
the  machine? 

118.  A  lineshaft  runs  250  R.  P.  M.     A  grinder  with  a  6  in.  pulley  is  to 
run  1550  R.  P.  M.     Determine  size  of  pulley  to  put  on  the  lineshaft  to  run 
the  grinder  at  the  desired  speed. 

119.  An  apprentice  was  given  100  bolts  to  thread.     He  completed  three- 
fifths  of  this  number  in  45  minutes  and  then  the  order  was  increased  so 
that  it  took  him  2  hours  for  the  entire  lot.     How  many  bolts  did  he  thread? 

120.  A  42  in.  planer  has  a  cutting  speed  of  30  ft.  per  minute  and  the  ratio 
of  cutting  speed  to  return  speed  of  the  table  is  1 : 2.8.     What  is  the  return 
speed  in  feet  per  minute? 

CHAPTER  IX 
PULLEY  AND  GEAR  TRAINS— CHANGE  GEARS 

60.  Direct  and  Inverse  Proportions. — A  proportion  formed  of 
numbers  of  castings  and  the  weights  of  metal  required  to  make 
them  is  a  direct  proportion,  because  the  amount  of  metal  required 
increases  directly  as  the  number  of  castings  increases. 


66  SHOP  ARITHMETIC 

When  two  pulleys  (or  gears)  are  running  together,  one  driving 
the  other,  the  larger  of  the  two  is  the  one  that  runs  the  slower. 
The  proportion  formed  from  their  diameters  and  revolutions  is, 
therefore,  called  an  Inverse  Proportion,  because  the  larger  pulley 
runs  at  the  slower  speed.  The  number  of  revolutions  of  one 
pulley  is  said  to  vary  inversely  as  its  diameter,  since  the  greater 
the  diameter,  the  less  the  number  of  revolutions  it  will  make. 

In  every  pair  of  gears  one  of  them  is  driving  the  other,  so  the 
one  can  be  called  the  driving  gear,  or  the  driver,  and  the  other  the 
driven  gear,  or  the  follower.  These  names  are  in  quite  general 
use  to  designate  the  gears  and  to  assist  in  keeping  the  propor- 
tions in  the  right  order.  Accordingly,  we  have  the  proportion: 


R.  P.  M.  of 
driven 


/R.  P.  M.  of\  =  /No.  of  teethN    /No.  of  teethN 
'  y      driver       /       \   on  driver   /  '  \  on  driven  / 


This  is  an  inverse  proportion  because  the  driver  and  the  driven 
are  in  the  reverse  order  in  the  second  ratio  from  what  they  are 
in  the  first  ratio.  Perhaps  this  can  be  seen  better  if  the  ratios  are 
written  as  fractions. 

R.  P.  M.  of  driven  _  No.  of  teeth  on  driver 
R.  P.  M.  of  driver       No.  of  teeth  on  driven 


Fio.  10. 

Here  the  reason  for  the  name  "inverse  proportion"  is  easily 
seen.  The  second  fraction  has  the  driver  and  the  driven  inverted 
from  what  they  are  in  the  first  fraction.  This  method  of  writing 
proportions  as  fractions  is  much  used  in  solving  problems  in 
gears  or  pulleys. 

61.  Gear  Trains. — A  gear  train  consists  of  any  number  of 
gears  used  to  transmit  motion  from  one  point  to  another.     Fig. 


PULLEY  AND  GEAR  TRAINS  67 

10  shows  the  simplest  form  of  gear  train,  having  but  two  gears. 
Fig.  1 1  shows  the  same  gears  A  and  J3,  as  in  Fig.  10,  but  with  a 
third  gear,  usually  called  an  intermediate  gear,  between  them. 
The  intermediate  gear  C  can  be  used  for  either  of  two  reasons: 

1.  To  connect  A  and  B  and  thus  permit  of  a  greater  distance 
between  the  centers  of  A  and  B  without  increasing  the  size  of 
the  gears;  or 

2.  To  reverse  the  direction  of  rotation  of  either  A  or  B.     If 
A  turns  in  a  clockwise  direction,  as  shown  in  both  Figs.  10  and 
11,  B  in  Fig.  10  will  turn  in  the  opposite,  or  counter-clockwise 
direction,  but  in  Fig.  11,  B  will  turn  in  the  same  direction  as  A. 


FIG.  11. 


The  introduction  of  the  intermediate  gear  C  has  no  effect  on 
the  speed  ratio  of  A  to  B.  If  A  has  48  teeth  and  B  96  teeth,  the 
speed  ratio  of  A  to  B  will  be  2  to  1  in  either  Fig.  10  or  Fig.  11. 

In  Fig.  10  suppose  A  to  be  the  driver. 

R.  P.  M.  of  driver  _  No.  of  teeth  on  driven 
R.  P.  M.  of  driven"  No.  of  teeth  on  driver 

R.  P.  M.  of  driver _  96 _2 
R.  P.  M.  of  driven"  48~I 

Hence,  the  speed  ratio  of  A  to  B  is  2  to  1. 

In  the  case  shown  in  Fig.  11  when  A  moves  a  distance  of  one 
tooth,  the  same  amount  of  motion  will  be  given  to  C,  and  C  must 
at  the  same  time  move  B  one  tooth.  To  move  B  96  teeth,  or 
one  revolution,  will  require  a  motion  of  96  teeth  on  A,  or  two 
revolutions  of  A.  Hence,  A  will  turn  twice  to  each  one  turn  of 

6 


68  SHOP  ARITHMETIC 

B,  or  the  speed  ratio  of  A  to  B  is  2  to  1,  just  as  in  the  case  of 
Fig.  10. 

62.  Compound  Gear  and  Pulley  Trains. — Quite  often  it  is 
desired  to  make  such  a  great  change  in  speed  that  it  is  practically 
necessary  to  use  two  or  more  pairs  of  gears  or  pulleys  to  accom- 
plish it!  If  a  great  increase  or  reduction  of  speed  is  made  by  a 
single  pair  of  gears  or  pulleys,  it  means  that  the  difference  in  the 
diameters  will  have  to  be  very  great.  The  belt  drive  of  a  lathe  is 
an  example  of  a  compound  train  of  pulleys,  though  here  the 
train  is  used  chiefly  for  other  reasons.  In  the  first  step,  the 
pulley  on  the  lineshaft  drives  a  pulley  on  the  countershaft;  then 
another  pulley  on  the  countershaft  drives  the  lathe.  The  back 
gearing  on  a  lathe  is  an  example  of  compound  gearing,  two 
pairs  of  gears  being  used  to  make  the  speed  reduction  from  the 
cone  pulley  to  the  spindle  and  face  plate. 


FIG.  12. 


Fig.  12  shows  a  common  arrangement  of  compound  gearing. 
Here  A  drives  B  and  causes  a  certain  reduction  of  speed.  B 
and  C  are  fastened  together  and  therefore  travel  at  the  same 
speed.  A  further  reduction  in  speed  is  made  by  the  two  gears 
C  and  D.  A  and  C  are  the  driving  gears  of  the  two  pairs  and 
B  and  D  are  the  driven  gears. 

In  making  calculations  dealing  with  compound  gear  or  pulley 
trains,  we  might  make  the  calculations  for  each  pair  as  explained 
in  Chapter  VIII  and  then  proceed  to  the  next  pair,  etc.,  but  this 
can  be  shortened  to  form  a  much  simpler  process. 

The  speed  ratio  for  a  pulley  or  gear  train  is  equal  to  the  product 
of  the  ratios  of  all  the  separate  pairs  of  pulleys  or  gears  making  up 
the  train. 


PULLEY  AND  GEAR  TRAINS  69 

In  using  this  principle  for  calculations,  the  ratios  are  written 
as  fractions  and  Ave  have  the  following  formula: 

R.  P.  M.  of  last  driven  gear    Product  of  Nos.  of  teeth  of  all  drivers 

K.  P.  M.  of  first  driver  1  roduct  of  Nos.  of  teeth  of  all  driven  gears 

Or,  if  we  want  the  ratio  stated  the  other  way  around  — 

R.  P.  M.  of  first  driver  Product  of  Nos.  of  teeth  of  all  driven  gears 

R.  P.  M.  of  last  driven  gear"  Product  of  Xos.  of  teeth  of  all  drivers 

Example  : 

Let  us  calculate  the  speed  ratio  for  the  train  of  gears  in  Fig.  12. 
This  would  be  the  ratio  of  the  speed  of  A  to  the  speed  of  D. 

A  is  the  first  driver  and  D  the  last  driven  gear,  and  the  ratio  of  their  speeds 
is  the  ratio  for  the  whole  train. 

Speed  of  A  ^  Teeth  on  Bx  Teeth  on  D 
Speed  of~Z>  ""  Teeth  on  A  X  Teeth  on  C 

2 
5       I 

10 


Hence, 

Speed  of  A:  Speed  of  D  =  10:l 
In  other  words,  A  revolves  10  times  as  fast  as  D. 

Problems  in  getting  the  speed  ratios  of  pulley  trains  are  solved 
in  the  same  way  except  that  diameters  are  used  instead  of  num- 
bers of  teeth. 

Speed  of  last  driven  pulley  _  Product  of  diameters  of  all  driving  pulleys 
Speed  of  first  driving  pulley     Product  of  diameters  of  all  driven  pulleys 

Example  : 

Let  us  take  the  pulley  train  of  Fig.  13  and  calculate  the  ratio  of 
the  speeds  of  pulleys  A  and  D.  A  and  C  are  the  drivers  and  B  and  D  are  the 
driven  pulleys. 

Speed  of  A  _  Diameter  of  B  X  Diameter  of  D 
Speed  of  D     Diameter  of  A  X  Diameter  of  C 
5 

_  5  __  i_ 

~36     7.2 
3     12 
Hence, 

Speed  of  A  :  Speed  of  D  =  1  :  7.2 

Trains  are  frequently  used  having  combinations  of  pulleys  and 
gears.  In  nearly  all  machine  tools,  we  will  find  both  pulleys  and 
gears  between  the  lineshaft  and  the  work.  In  wood-working 


70  SHOP  ARITHMETIC 

machinery,  on  the  other  hand,  we  usually  find  only  pulleys  and 
belts,  on  account  of  the  high  speeds  at  which  the  machines  are 
run.  In  calculating  the  speed  ratios  of  these  combined  trains, 
we  can  use  the  diameters  of  the  pulleys  and  the  numbers  of 
teeth  of  the  gears  in  the  same  formula. 

/R.  P.  M.  of\  /Product  of  diameters\  x  /Product  of  teeth  oA 
\first  driver/  _  \of  all  driven  pulleys/  \  all  driven  gears  / 
/R.  P.  M.  of\  =  /Product  of  diametersX  /Product  of  teeth  of\ 

yast  driven  j       \of  all  driving  pulleys  j  x  \^    all  driving  gears  ) 


Fio.  13. 

If  a  problem  calls  for  the  calculation  of  the  size  of  one  pulley 
or  gear  in  a  train,  all  the  others  and  the  speed  ratio  being  known, 
start  at  one  or  both  ends  of  the  train  and  work  toward  the  gear  or 
pulley  in  question  until  you  get  a  proportion  which  will  give  the 
desired  quantity. 

Example: 

The  punch  shown  m  Fig.  14  is  to  be  set  up  so  that  it  will  make  20 
strokes  per  minute.  (The  80  tooth  gear  must,  therefore,  run  20  R.  P.  M.) 
The  punch  is  to  be  driven  from  a  countershaft  and  we  want  to  calculate  the 
size  of  the  pulley  to  put  on  the  countershaft,  to  drive  the  punch  at  the  desired 
speed.  We  find  that  the  main  lineshaft  runs  240  R.  P.  M.  and  carries  a  16-in. 
pulley  which  drives  a  24-in.  pulley  on  the  countershaft. 

Working  from  the  lineshaft: 

24  in.xR.  P.  M.  of  countershaft  =  16X240. 


R.  P.  M.  of  countershaft=^p  =  160  R.  P.  M. 
24 


PULLEY  AND  GEAR  TRAINS 

Working  from  the  punch: 

20XR.  P.  M.  of  20  T  gear  =  80X20. 
R.  P.  M.  of  20  T  gear  =  -^--  =  80 


71 


This  is  also  the  R.  P.  M.  of  the  24-in.  pulley  and  this  pulley  is  driven 
by  the  unknown  pulley  on  the  countershaft,  which  we  have  found 
runs  160  R.  P.  M. 
IGOxDiam.  of  pulley  =  80X24 

1920 
Diam.  of  pulley  =  -^r-  =  12  in.,  Answer. 


Fio.  1 1. 


Another  way  to  solve  this  would  be  to  write  put  an  equation  for  the  entire 
train,  using  X  to  represent  the  pulley  whose  size  we  want  to  find. 

20  =  16  XXX  20 
240  "24X24X80 

?  1 


== 

12     6X24     6     24 

XI  IX  1 

Now,  if  ^  X  „  is  to  equal  ^-.y  then  —  must  be  0«  which  would  be  when  A" 


12 


24 


is  12. 

Hence,  X  — 12  in.,  Answer. 


72  SHOP  ARITHMETIC 

63.  Screw  Cutting. — Most  lathes  are  equipped  with  a  small 
plate  giving  the  necessary  gears  to  use  for  cutting  different 
threads,  but  every  good  machinist  should  know  how  to  calculate 
the  proper  gear  setting  for  such  work.  This  is  a  simple  problem 
in  gear  trains  and  should  cause  no  difficulty  for  the  man  who 
understands  the  principles  of  gear  trains. 

The  lathe  carriage  and  tool  are  moved  by  a  "lead  screw" 
having  usually  2,  4,  6,  or  8  threads  per  inch.  If  a  lathe  has  a 
lead  screw  having  6  threads  per  inch,  each  revolution  of  the  lead 
screw  will  move  the  carriage  £  in. ;  a  4  pitch  screw  would  move 
the  carriage  \  in.  for  each  revolution  of  the  screw.  Then,  if  the 
spindle  of  the  lathe  and  the  lead  screw  turn  at  the  same  speed,  the 
lathe  will  cut  a  thread  of  the  same  pitch  as  that  on  the  lead  screw. 
If  a  finer  thread  is  wanted  than  that  on  the  lead  screw,  the 
spindle  should  make  more  turns  than  does  the  lead  screw. 
Suppose  we  want  to  cut  24  threads  per  inch  and  have  a  6  thread 
per  inch  lead  screw.  It  will  require  6  turns  of  the  lead  screw  to 
move  the  carriage  1  inch.  Meanwhile,  the  work  should  revolve 
24  times.  Then  the  ratio  of  spindle  speed  to  lead  screw  speed 
should  be  4:1 

Speed  of  spindle  Threads  per  inch  to  be  cut 

Speed  of  lead  screw     Threads  per  inch  on  lead  screw 

The  first  driving  gear  is  that  on  the  spindle,  while  the  last  driven 
gear  is  that  on  the  end  of  the  lead  screw.     Hence, 

Threads  per  inch  to  be  cut    _  Product  of  Nos.  of  teeth  on  driven  gears 
Threads  per  inch  on  lead  screw     Product  of  Nos.  of  teeth  on  driving  gears 

PROBLEMS 

121.  In  Fig.  10,  if  we  removed  the  48  tooth  gear  and  put  a  64  tooth  gear 
in  its  place,  what  would  be  the  speed  ratio  of  A  to  B? 

122.  In  Fig.  11,  if  B  makes  6  revolutions,  how  many  turns  will  C  make 
and  how  many  will  A  make? 

123.  What  would  be  the  speed  ratio  of  the  train  of  Fig.  12  if  we  put  a 
32T  gear  on  at  C  and  a  48T  gear  at  D? 

124.  The  lineshaft  in  Fig.  15  runs  250  R.  P.  M.     Determine  the  size  of 
lineshaft  pulley  to  run  the  grinder  at  1550  R.  P.  M.  using  the  countershaft 
as  shown  in  the  figure. 

126.  A  machinist  wishes  to  thread  a  pipe  on  a  lathe  having  2  threads  per 
inch  on  the  lead  screw.  There  are  to  be  11^  threads  per  inch  on  the  pipe. 
What  is  the  ratio  of  the  speeds  of  the  spindle  and  the  lead  screw? 

126.  Two  gears  are  to  have  a  speed  ratio  of  4.6  to  1.  If  the  smaller  gear 
has  15  teeth,  what  must  be  the  number  of  teeth  on  the  larger  gear? 


PULLEY  AND  GEAR  TRAINS 


73 


COUNT  CJ?SHAFT 


GRINDLR 


FIQ.  15. 


FIG.  IS. 


74  SHOP  ARITHMETIC 

127.  It  has  been  decided  to  equip  the  punch  in  Fig.  14  with  a  motor  drive 
by  replacing  the  fly  wheel  with  a  large  gear  to  be  driven  by  a  small  pinion 
on  the  motor.     If  the  motor  runs  800  R.  P.  M.,  and  has  a  16  tooth  pinion, 
what  must  be  the  number  of  teeth  on  the  other  gear?     Speed  of  the  punch 
to  be  20  strokes  per  minute. 

128.  A  street  car  is  driven  through  a  single  pair  of  gears,  a  large  gear  on 
the  axle  being  driven  by  a  smaller  one  on  the  motor  shaft.     If  a  car  has 
33-in.  wheels  and  a  gear  ratio  of  1 :4,  how  fast  would  the  car  go  when  the 
motor  is  running  1200  R.  P.  M.? 

129.  Fig.  16  shows  the  head  stock  for  a  lathe.     The  cone  pulley  carries 
with  it  the  cone  pinion  A,  which  drives  the  back  gear  B.     B  is  connected 
solidly  with  the  back  pinion  C  which  drives  the  face  gear  D.     If  the  gears 
have  the  following  numbers  of  teeth,  determine  the  back  gear  ratio  (speed 
of  A :  speed  of  D) : 

Teeth  on  cone  pinion,  A  28 
Teeth  on  back  gear,  B  82 
Teeth  on  back  pinion,  C  25 
Teeth  on  face  gear,  D  74 

130.  If  you  were  to  cut  a  20  pitch  thread  on  a  lathe  having  a  4  pitch  lead 
screw,  what  would  be  the  ratio  of  the  speeds  of  the  spindle  and  the  lead 
screw? 


CHAPTER  X 
AREAS  AND  VOLUMES  OF  SIMPLE  FIGURES 

64.  Squares. — In  taking  up  the  calculation  of  areas  of  surfaces 
and  the  volumes  and  weights  of  objects,  the  expressions  "square" 
and  "square  root"  will  be  met  and  must  be  understood.     To 
one  unfamiliar  with  these  names  and  the  corresponding  operations 
the  signs  and  operations  themselves  seem  difficult.     They  are  in 
reality  very  simple.     The  square  of  a  number  is  simply  the  prod- 
uct of  the  number  multiplied  by  itself;  the  square  of  2  is  2  X2  =  4; 
the  square  of  5  is  5X5  =  25;  the  square  of  12.5  is  12.5X12.5  = 
156.25.     Instead  of  writing  2x2  or  5X5,  it  is  customary   to 
write  22  and  52.     These  are  read  "2  squared"  and  "5  squared." 
12.52=12.5    squared,  and  so  on.     The  little  2  at  the    upper 
right  hand  corner  is  called  the  Exponent. 

65.  Square  Root. — The  square  root  of  a  given  number  is 
simply  another  number  which,  when  multiplied  by  itself  (or 
squared),  produces  the  given  number.     Thus,  the  square  root 
of  4  is  2,  since  2  multiplied  by  itself  (2X2)  gives  4.     The  square 
root  of  9  is  3,  since  3X3=32  =  9.     Square  root  is  the  reverse  of 
square,  so  if  the  square  of  5  is  25  the  square  root  of  25  is  5.     The 
mathematical  sign  of  square  root,  called  the  radical  sign,  is  \/7 
Then  V9  =  3;  V  25  =  5.     These  expressions  are  read  "the  square 
root  of  9=3";  "the  square  root  of  25  =  5".     Square  roots  of 
larger  numbers  can  usually  be  found  in  handbooks  and  the 
actual  process  of  calculating  them,  which  is  somewhat  compli- 
cated, will  be  taken  up  later  on. 

66.  Cubes  and  Higher  Powers. — In  the  same  way  that  22 
(2  squared)  =2X2 =4,   23  (2  cubed)  =2x2x2  =  8.   The  exponent 
simply  indicates  how  many  times  the  number  is  used  as  a  factor, 
or  how  many  times  it  is  multiplied  together.     4s  =  4x4x4  =  64. 
33=3X3X3  =  27. 

Just  as  square  root  is  the  reverse  of  square,  so  cube  root  is 

the  reverse  of  cube.     The  sign  for  cube  root  is  V7    So  if  38  = 

3X3X3=27,  then   ^27  =  3.     Sometimes  a  factor  is  repeated 

more  than  3  times,  in  which  case,  the  exponent  indicates  the 

7  75 


76 


SHOP  ARITHMETIC 


number  of  times.  2*  means  2x2x2x2  and  is  read  "2  to  the 
fourth  power."  2s  =  2X2X2X2X2  and  is  read  "2  to  the  fifth 
power,"  and  so  on.'"  The  roots  are  indicated  in  the  same  way. 
^16  =  fourth  root  of  16  =  2;  54  =  5X5X5X5  =  625,  etc. 

67.  Square  Measure. — Before  going  further,  it  will  be  well 
to  get  clearly  in  mind  just  what  the  term  "Square"  means  in 
terms  of  the  things  we  see.  Areas  of  figures  are  measured  in 
terms  of  the  "square"  unit.  For  instance,  if  the  dimensions  of 
the  base  of  a  milling  machine  are  3  ft.  by  5  ft.,  the  floor  space 
covered  by  this  base  is  15  square  feet.  In  this  case  the  area  is 


•5'- 


FIQ.  17. 

measured  by  the  unit  known  as  the  square  foot.  A  Square 
Foot  is  a  surface  bounded  by  a  square  having  each  side  1  ft. 
in  length.  In  case  of  the  milling  machine  base  represented  in 
Fig.  17,  there  are  by  actual  count  15  sq.  ft.  in  this  surface  and 
this  is  readily  seen  to  be  the  product  of  the  length  and  the  breadth 
of  the  base,  since  3X5  =  15. 

The  Square  Inch  is  another  common  unit  of  area.  This  is 
much  smaller  than  the  square  foot,  being  only  one-twelfth  as 
great  each  way.  If  a  square  foot  is  divided  into  square  inches 
it  will  be  seen  to  contain  12X12  or  144  sq.  in.  (see  Fig.  18).  It 
will  be  readily  seen  that  the  area  of  any  square  is  equal  to  the 
product  of  the  side  of  the  square  by  itself.  In  other  words,  the 
area  of  a  square  equals  the  side  "squared"  (referring  to  the 
process  explained  in  Article  64).  Looking  at  it  the  other  way 
around,  the  square  of  any  number  can  be  represented  by  the 
area  of  a  square  figure,  one  side  of  which  represents  the  number 
itself.  The  actual  things  which  the  number  represents  makes  no 
difference  whatever.  If  the  side  of  a  square  is  5  in.,  the  area  is 


AREAS  AND  VOLUMES  OF  SIMPLE  FIGURES     77 

25  sq.  in.;  if  the  side  is  5  ft.,  the  area  is  25  sq.  ft.  If  we  simply 
have  the  number  5,  its  square  is  25,  no  matter  what  kind  of 
things  the  5  may  refer  to. 

As  mentioned  before,  1  sq.  ft.  is  the  area  of  a  square  1  ft.  on 
each  side  and,  if  divided  into  square  inches,  will  be  found  to 
contain  122  or  144  sq.  in.  Likewise,  a  square  yard  is  3  ft.  on 
each  side  and,  therefore,  contains  3*  =  9  sq.  ft.  The  following 
table  gives  the  relation  between  the  units  ordinarily  used  in 
measuring  areas: 


FIG.  18. 


MEASURES  OF  AREA  (SQUARE  MEASURE) 

144  square  inches  (sq.  in.)  =  1  square  foot  (sq.  ft.) 

9  square  feet=l  square  yard  (sq.  yd.) 
30  J  square  yards  =  1  square  rod  (sq.  rd.) 
160  square  rods=l  acre  (A) 

640  acres  =  1  square  mile  (sq.  mi.) 

68.  Area  of  a  Circle. — If  a  circle  is  drawn  in  a  square  as  shown 
in  Fig.  19,  it  is  easily  seen  that  it  has  a  smaller  area  than  the 
square  because  the  corners  are  cut  off.  The  area  of  the  circle 
is  always  a  definite  part  of  the  area  of  the  square  drawn  on  its 
diameter,  the  area  of  the  circle  being  always  .7854  times  the 
area  of  the  square.  This  number  .7854  happens  to  be  just  one- 
fourth  of  the  number  3.1416  given  in  Chapter  VII.  Just  why 
this  is  so  will  be  shown  later  on.  If  the  diameter  of  the  circle 
=  10  in.,  as  in  Fig.  19,  the  area  of  the  square  is  100  sq.  in.  and  the 
area  of  the  circle  is  .7854x100  =  78.54  sq.  in.  You  can  prove 
this  to  your  own  satisfaction  in  the  following  manner.  Cut  a 


78 


SHOP  ARITHMETIC 


square  of  carHboard  of  any  size,  and  from  the  center  describe  a 
circle  as  shown  just  touching  on  all  four  sides.  Weigh  the  square, 
and  then  cut  out  the  circle  and  weigh  it.  The  circle  will  weigh 
. 7854  X  weight  of  the  square.  A  pair  of  balances  such  as  are 
found  in  a  drug  store  are  the  best  for  this  experiment. 


10"- 


FIQ.  19. 


Rule  for  Area  of  Circle.  —  The  area  of  any  circle  is  obtained  by 
squaring  the  diameter  and  then  multiplying  this  result  .by  .7854. 
If  written  as  a  formula  this  rule  would  read 


where  A  =  area  of  a  circle 
of  which  D  =  the  diameter. 

Example  : 

Find  the  area  of  a  circle  3  in.  in  diameter. 
A=   .7854  XD2 
A=   .7854  X32 

=   .7854X9  _ 

=  7.0686  sq.  in.,  Answer. 

If  you  think  a  little  you  will  see  that,  if  the  diameter  is  doubled, 
the  area  is  increased  four  times.  This  can  also  be  seen  from  Fig. 
20.  The  diameter  of  the  large  circle  is  twice  that  of  one  of  the 
small  circles,  but  its  area  is  four  times  that  of  one  of  the  small 
circles.  This  is  a  very  important  and  useful  law  and  may  be 
stated  as  follows:  "The  areas  of  similar  figures  are  to  each  other 
as  the  squares  of  their  like  dimensions."  A  2  in.  circle  contains 
22X.  7854  =  3.  1416  sq.  in.,  while  a  6  in.  circle  contains  62X 
.7854  =  28.2744  sq.  in.,  or  nine  times  as  much.  This  we  can  find 


AREAS  AND  VOLUMES  OF  SIMPLE  FIGURES     79 

by  saying  the  6  in.  circle  has  three  times  the  diameter  of  the 
2  in.  circle  and,  therefore,  the  area  is  32,  or  nine  times  as  great. 
A  piece  of  steel  plate  6  in.  in  diameter  weighs  nine  times  as  much 
as  a  piece  2  in.  in  diameter  of  the  same  thickness.  Likewise  a 
10  in.  square  has  four  times  the  area  of  a  5  in.  square.  If  we 


Fio.  20. 


let  A  represent  the  area  of  the  larger  circle,  a  the  area  of  the 
smaller  circle,  D  the  diameter  of  the  larger  circle,  and  d  the  diam- 
eter of  the  smaller  circle,  then  we  have  the  direct  proportion: 


69.  The  Rectangle.  —  When   a   four-sided   figure   has   square 
corners  it  is  a  Rectangle.     Each  side  of  a  brick  is  a  rectangle. 


3" 


Fio.  21. 


A  Square  is  a  special  kind  of  rectangle  having  all  the  sides  equal. 
The  area  of  a  rectangle  is  obtained  by  multiplying  the  length  by 
the  breadth.  In  Fig.  21  the  area  is  2X3  =  6  sq.  in.,  as  can  be 
seen  by  counting  the  1-in.  squares,  which  each  contain  1  sq.  in. 


80 


SHOP  ARITHMETIC 


70.  The  Cube. — Just,  as  the  square  of  a  number  is  represented 
by  the  area  of  a  square,  one  side  of  which  represents  the  number, 

so  the  cube  of  a  number  is  represented 
by  the  volume  of  a  cubical  block,  each 
edge  of  which  represents  the  number. 
The  volume  of  a  cube  which  is  10  in. 
on  each  edge  is  10  X 10  X 10  - 1000  cubic 
inches,  and  since. this  is  obtained  by 
"cubing"  10  (103  =  10X10X10  =  1000), 
we  can  see  that  the  cube  of  a  number 
FIQ  22  can  be  represented  by  the  volume  of  a 

cube,  the  edge  of  which  represents  the 

number.  If  the  edge  of  the  cube  is  one-half  as  long,  that  is  5  in., 
the  volume  is  5X5X5  =  1 25  cubic  inches,  or  only  £  the  volume  of 
the  10  in.  cube. 


^s>^> 

I 

,0"p\ 

0 

1 

1 

\ 

^ 

Fia.  23. 


MEASURES  OF  VOLUME  (CUBICAL  MEASURE) 

1728  cubic  inches  (cu.  in.)  =  1  cubic  foot  (cu.  ft.) 
27  cubic  feet  =  l  cubic  yard  (cu.  yd.) 
(Larger  units  than  cubic  yards  are  seldom,  if  ever,  used.) 

71.  Volumes  of  Straight  Bars. — A  piece  1  in.  long  cut  from  a  bar 
will  naturally  contain  just  as  many  cu.  in.  as  there  are  sq.  in. 
on  the  end  of  the  bar.  In  the  billet  shown  in  Fig.  24,  there  are 
3x4  =  12  sq.  in.  on  the  end  of  the  bar,  and  a  piece  1  in.  long 
contains  12  cu.  in.  The  entire  billet,  contains  10  slices  just  like 
this  one,  so  there  are  12X10  =  120  cu.  in.  in  the  entire  billet. 


AREAS  AND  VOLUMES  OF  SIMPLE  FIGURES    81 


Therefore,  we  see  that  to  find  the  number  of  cu.  in.  in  any 
straight  bar  we  proceed  as  follows: 

Calculate  the  area  of  one  end  of  the  bar  in  square  inches; 
then  multiply  this  result  by  the  length  of  the  bar  in  inches; 
the  result  will  be  the  number  of  cubic  inches  in  the  bar.  For 
bars  of  square  or  rectangular  section,  the  volume  is  the  product 
of  the  three  dimensions,  length,  breadth,  and  thickness.  If  L, 
B,  and  T  represent  the  length,  breadth,  and  thickness,  and  V 
stands  for  the  volume,  then 

V=LXBXT 


L 


10"- 


Fio.  24. 


Example : 


How  many  cubic  inches  of  steel  in  a  bar  2  in.  square  and  4  ft.  long? 
4  ft.  =48  in. 
V=LXBXT 
=  48X2X2  =  192  cu.  in.,  Answer. 

For  round  bars,  the  area  on  the  end  is  .7854  times  the  square  of 
the  diameter,  and  this,  multiplied  by  the  length,  gives  the  volume. 
Then,  if  D  represents  the  diameter  of  the  bar  and'L  its  length, 
the  volume  V  will  be 

7=.7854X£2XL 

This  will  apply  equally  well  to  thin  circular  plates  or  to  long 
bars  or  shafting.  With  thin  plates,  we  would  naturally  speak 
of  thickness  (T)  instead  of  length  (L).  Fig.  25  shows  that  the 
two  objects  have  the  same  shape  except  that  their  proportions 
are  different. 

Examples : 

1.  How  many  cubic  inches  of  steel  in  a  shaft  2  in.  in  diameter  and  12  ft. 
long? 

12  ft.  =  12X12  =  144  in.,  length  of  bar. 
F  =  .7854XD2XL 
V  =  .  7854  X22X 144 
V  =  .7854  X  4  X 144  =452.39  cu.  in.,  Answer. 


82 

2.  How  many  cubic  inches  in  a  blank  for  a  boiler  head  60  in.  in  diameter 
and  TTT  in.  thick? 


7  =  .7854X602X  ^ 

F  =  . 7854 X 3600 XTF  =  1237  cu.  in.,  Answer. 


Fio.  25. 


72.  Weights  of  Metals. — The  chief  uses  in  the  shop  for  calcu- 
lations of  volume  are  in  finding  the  amount  of  material  needed 
to  make  some  object;  in  finding  the  weight  of  some  object  that 
cannot  be  conveniently  weighed;  or  in  finding  the  capacity  of 
some  bin  or  other  receptacle.  Having  obtained  the  volume  of 
an  object,  it  is  only  necessary  to  multiply  the  volume  by  the 
known  weight  of  a  unit  volume  of  the  material  to  get  the  weight 
of  the  object.  In  the  case  of  the  shaft  of  which  we  just  got  the 
volume,  1  cu.  in.  will  weigh  about  .283  lb.,  so  the  total  weight 
of  the  shaft  will  be 

452 . 39  X  .  283  =  128 . 0  +  pounds. 
The  weight  of  the  boiler  head  will  be 

1237  X  .  283  =  350  +  pounds. 

The  following  table  gives  the  weights  per  cubic  inch  and  per 
cubic  foot  for  the  most  common  metals  and  also  for  water: 


Material 

.  1  cu.  in. 

1  cu.  ft. 

Cast  iron  

.260  Ib. 

450  Ib. 

Wrought  iron  

.278  Ib. 

480  Ib. 

Steel  

.283  Ib. 

489  Ib. 

Brass  

.  301  Ib. 

520  Ib. 

Copper.  . 

.3181b. 

550  Ib. 

Lead  

.411  Ib. 

711  Ib. 

Aluminum  

.  094  Ib. 

162  Ib. 

Water  

.036  Ib. 

62.4  Ib. 

73.  Short  Rule  for  Plates. — A  flat  wrought  iron  plate  £  in. 
thick  and  1  ft.  square  will  weigh  5  Ib.,  since  12X12x^  =  18  cu. 
in.,  and  18 X. 278  =  5  Ib.     The  rule  obtained  from  this  is  very 
easy  to  remember  and  is  very  useful  for  plates  that  have  their 
dimensions  in  exact  feet. 

Rule. — Weight  of  flat  iron  plates  =  area  in  square  feetX  number 
of  eighths  of  an  inch  in  thickness  X  5.  This  rule  can  also  be  used 
for  steel  plates  by  adding  2  per  cent,  to  the  result  calculated  from 
the  above  rule. 

Example : 

3 

Find  the  weight  of  a  steel  plate  30  in.  X96  in.  X  s  in. 

o 

1  3 

30  in.  =  2^  ft.,  96  in.  =-  8  ft.,  g  in.  =  3  eighths. 

2^X8X3X5  =  300  (weight  if  it  were  of  wrought  iron). 

2%  of  300  =  6  Ib. 

300  +  6  =  306  Ib.;  weight  of  steel  plate,  Answer. 

If  this  weight  is  calculated  by  first  getting  the  cubic  inches  of  steel,  we  get : 

Q 

30  X  96  X 77  =  1080  cu.  in. 

o 

1080 X. 283  =  305.64  Ib.  weight  of  steel  plate,  Answer. 

We  see  that  the  results  check  as  closely  as  could  be  expected 
and,  in  fact,  different  plates  of  supposedly  the  same  size  would 
differ  as  much  as  this  because  of  differences  in  rolling. 

74.  Weight  of  Casting  from  Pattern. — In  foundry  work,  it  is 
often  desired  to  get  the  approximate  weight  of  a  casting  in  order 
to  calculate  the  amount  of  metal  needed  to  make  it.     The  prob- 


84 


SHOP  ARITHMETIC 


able  weight  of  the  casting  can  be  obtained  closely  enough  by 
.weighing  the  pattern  and  multiplying  this  weight  by  the  proper 
number  from  the  following-  table.  In  case  the  pattern  contains 
core  prints,  the  weight  of  these  prints  should  be  calculated  and 
subtracted  from  the  pattern  weight  before  multiplying;  or  else  the 
total  pattern  weight  can  be  multiplied  first  and  then  the  weight 
of  metal  which  would  occupy  the  same  volume  as  the  core  print 
be  subtracted  from  it. 


PROPORTIONATE  WEIGHT  OF  CASTINGS  TO  WEIGHT  OF 
WOOD  PATTERNS 


For  each  1  Ib.  weight  of 
pattern    when    made    of 
(less  weight  of  core  prints) 

Casting  will  weigh  if  made  of 

Cast  iron 

Copper  or 
bronze 

Brass 

Aluminum 

White  pine  

16. 
12. 
10.2 
10.6 
0.84 
2.6 
0.95 

19.6 
14.7 
12.5 
13. 
1. 
3.2 
1.3 

18.5 
14. 
11.7 
12.3 
0.95 
3.1 
1.2 

5.9 
4.5 
3.8 
3.9 
0.32 
0.95 
0.38 

Mahogany  

Pear  wood  

Birch        

Brass  

Aluminum     

Cast  iron  

PROBLEMS 

131.  Find  the  weight  of  a  piece  of  steel  shafting  2  in.  in  diameter  and 
20  ft.  long. 

132 .  What  is  the  weight  of  a  billet  of  wrought  iron  4  in.  square  and  2  ft. 
8  in.  long? 

133.  What  would  a  steel  boiler  plate  36  in.  by  108  in.  by  \  in.  weigh? 

134.  A  cast  steel  cylinder  is  42  in.  inside  diameter,  4  ft.  6  in.  long  and 
1J  in.  thick.     Find  its  weight. 

135.  A  steam  engine  cylinder  4£  in.  inside  diameter  has  the  cylinder 
head  held  on  by  four  studs.     When  the  pressure  in  the  cylinder  is  125  Ib. 
per  square  inch,  what  is  the  total  pressure  on  the  cylinder  head  and  what 
is  the  pull  in  each  stud? 

136.  50  studs  1\  in.  long  and  1  in.  in  diameter  are  to  be  cut  from  cold 
rolled  steel.     Find  the  length  and  weight  of  bar  necessary,  allowing  \  in. 
per  stud  for  cutting  off. 

137.  What  would  be  the  weight  of  a  \  in.  by  3  in.  wagon  tire  for  a  40  in. 
wheel?     (Length  of  stock  =  circumference  of  &  39 J  in.  circle.) 


SQUARE  ROOT 


85 


138.  A  copper  billet  2  in.  by  8  in.  by  24  in.  is  rolled  out  into  a  plate  of 
No.  10  B.  &  S.  gage.     The  thickness  of  this  gage  is  .1019  in.     What  would 
be  the  probable  area  of  this  plate  in  square  feet? 

139.  The  steel  link  shown  in  Fig.  26  is  made  of  f  in.  round  steel  (round 
steel  f  in.  in  diameter).    Find  the  length  of  bar  necessary  to  make  it  and 
then  find  the  weight  of  the  link. 

140.  A  steel  piece  is  to  be  finished  as  shown  in  the  sketch  below  (Fig.  27). 
The  only  stock  available  from  which  to  make  it  is  4  in.  in  diameter.     Com- 
pute the  length  of  the  4  in.  stock  which  must  be  upset  to  make  the  piece  and 
have  -5*5  extra  stock  all  over  for  finishing. 


T 


Fio.  26. 


Fio.  27. 


CHAPTER  XI 

• 

SQUARE  ROOT 

76.  The  Meaning  of  Square  Root. — The  previous  chapter 
showed  the  usefulness  of  squares  in  finding  areas  and  of  cubes  in 
finding  volumes.  Problems  often  arise  in  which  it  is  necessary 
to  find  one  edge  of  a  square  or  cube  of  which  only  the  area  or 
,  volume  is  given.  For  instance,  what  must  be  the  side  of  a  square 
so  that  its  area  will  be  9  sq.  in.?  The  length  of  the  side  must  be 
such  that  when  multiplied  by  itself  it  will  give  9  sq.  in.  A 
moment's  thought  shows  that  3X3  =  9,  or  32  =  9.  Therefore, 
3  is  the  necessary  side  of  the  square.  Finding  such  a  value  is 
called  Extracting  the  Square  Root,  and  is  represented  by  the  sign 
V  called  the  square  root  sign  or  radical  sign.  Thus  v9  =  3 ; 
\/16  =  4.  To  make  clear  the  idea  of  extracting  square  roots, 
the  student  should  consider  it  as  the  reverse  or  "the  undoing" 


86  SHOP  ARITHMETIC 

of  squaring,  just  as  division  is  the  reverse  of  multiplication  or  as 
subtraction  is  the  reverse  of  addition. 

52  =  25,  and  its  reverse  is:  \/25  =  5. 

The  square  roots  of  some  numbers,  like  4,  9,  16,  25,  36,  49, 
64,  81,  etc.',  are  easily  seen,  but  we  must  have  some  method  that 
will  apply  to  any  number.  There  are  several  methods  of  finding 
square  root,  of  which  two  are  open  to  the  student  of  shop  arith- 
metic: (1)  by  actual  calculation;  (2)  by  the  use  of  a  table  of 
squares  or  square  roots.  A  third  method  which  uses  logarithms 
will  be  explained  in  the  chapters  on  logarithms.  In  many 
handbooks  will  be  found  tables  giving  the  square  roots  of 
numbers,  but  we  must  learn  some  method  that  can  be  used 
when  a  table  is  not  available  and  the  method  that  will  now 
be  explained  should  be  used  throughout  the  work  in  this 
chapter. 

76.  Extracting  the  Square  Root. — The  first  step  in  finding  the 
square  root  of  any  number  is  to  find  how  many  figures  there  are 
in  the  root.  This  is  done  by  pointing  off  the  number  into 
periods  or  groups  of  two  figures  each,  beginning  at  the  decimal 
point  and  working  each  way. 

12  =  1  102=1'00  1002=1'00'00 

From  these  it  is  evident  that  the  number  of  periods  indicates 
the  number  of  figures  in  the  root.  Thus  the  square  root  of 
103684  contains  3  figures  because  this  number  (10'36'84)  con- 
tains three  periods.  Also  the  square  root  of  6'50'25  contains 
three  figures  since  there  are  three  periods.  (The  extreme  left 
hand  period  may  have  1  or  2  figures  in  it.)  We  must  not  forget 
that,  for  any  number  not  containing  a  decimal,  a  decimal  point 
may  be  placed  at  the  extreme  right  of  the  number.  Thus  the 
decimal  point  for  62025  would  be  placed  at  the  right  of  the  number 
(as  62025.) 

The  method  of  finding  the  square  root  of  a  number  can  best 
be  explained  by  working  some  examples  and  explaining  the  work 
as  we  go  along.  The  student  should  take  a  pencil  and  a  piece 
of  paper  and  go  through  the  work,  one  step  at  a  time,  as  he  reads 
the  explanation. 

Example : 

Find  the  square  root  of  186624. 

Point  off  into  periods  of  two  figures  each  (18'66'24)  and  it  will  be  seen 
that  there  are  3  figures  in  the  root.  The  work  is  arranged  very  similarly  to 
division. 


SQUARE  ROOT  87 

18'66'24(432  Explanation:   First  find  the  largest 

number  whose  square  is  equal  to  or 


2X40  =  80 
JJ 
83 

2X430  =  860 

2 

862 


9~fiS~  less  ^an  I**,  *ke  ^rst  Pefiod.     This  is 

4,  since  52  is  more  than  18.  Write  the 
4  to  the  right  for  the  first  figure  of  the 
root  just  as  the  quotient  is  put  down 


17  24  in  long  division.     The  first  figure  of 

the  root  is  4.     Square  the  4  and  write 
17  24  its  square  (16)  under  the  first  period 


(18)  and  subtract,  leaving  2. 

Bring  down  the  next  period  (66)  and  annex  it  to  the  remainder,  giving 
266  for  what  is  called  the  dividend.  Annex  a  cipher  to  the  part  of  the  root 
already  found  (4)  giving  40;  then  multiply  this  by  2,  making  80,  which  is 
called  the  trial  divisor.  Set  this  off  to  the  left.  Divide  the  dividend  (266) 
by  the  trial  divisor  (80).  We  obtain  3,  which  is  probably  the  next  figure  of 
the  root.  Write  this  3  in  the  root  as  the  second  figure  and  also  add  it  to  the 
trial  divisor,  giving  83,  which  is  the  final  divisor.  Multiply  this  by  the  figure 
of  the  root  just  found  (3)  giving  249.  Subtract  this  from  the  dividend  (266) 
leaving  17. 

Bring  down  the  next  period  (24)  and  annex  to  the  17,  giving  a  new 
dividend  1724.  Repeat  the  preceding  process  as  follows:  Annex  a  cipher 
to  the  part  of  the  root  already  found  (43)  giving  430;  and  multiply  by  2, 
giving  860,  the. trial  divisor.  Divide  the  dividend  by  this  divisor  and  ob- 
tain 2  as  the  next  figure  of  the  root.  Put  this  down  as  the  third  figure  of 
the  root  and  also  add  it  to  the  trial  divisor,  giving  862  as  the  final  divisor. 
Multiply  this  by  the  2  and  obtain  1724,  which  leaves  no  remainder  when 
subtracted  from  the  dividend.  As  there  are  no  more  periods  in  the  original 
number,  the  root  is  complete. 

77.  Square  Roots  of  Mixed  Numbers. — If  it  is  required  to  find 
the  square  root  of  a  number  composed  of  a  whole  number  and 
a  decimal,  begin  at  the  decimal  point  and  point  off  periods  to 
right  and  left.     Then  find  the  root  as  before. 
Example : 

Find  the  square  root  of  257.8623 
2'57.86'q23'00(16.058+,  Answer. 


20 
_6 

26 


_1 to  or  less  than  2,  the  first  period.  Proceeding  as 

I  57  before,  we  get  6  for  the  second  figure.  After 

subtracting  the  second  time  (at  a)  we  find  that 
,  eg  the  trial  divisor  320  is  larger  than  the  dividend 
186.  In  this  case,  we  place  a  cipher  in  the  root, 

annex  another  cipher  to  320  making  3200,  annex 

the  next  period,  23,  to  the  dividend  and  then 
1  60  25  proceed  as  before.  If  the  root  proves,  as  in  this 

25  98  00  case,  to  be  an  interminable  decimal  (one  that 

does  not  end)  continue  for  two  or  three  decimal 


5 


320.5 


32100 

8 

32108 


25  68  64 


places  and  put  a  +  sign  after  the  root  as  in  divi- 
sion. In  this  example  the  decimal  point  comes 
29  36  after  16,  because  there  must  be  two  figures  in  the 
whole  number  part  of  the  root  since  there  are 
two  periods  in  the  whole  number  part  of  our 
original  number. 

78.  Square  Roots  of  Decimals. — Sometimes,  in  the  case  of  a 
decimal,  one  or  more  periods  are  composed  entirely  of  ciphers. 
The  root  will  then  contain  one  cipher  following  the  decimal  point 
for  each  full  period  of  ciphers  in  the  number. 


88  SHOP  ARITHMETIC 

Example : 

Take  .0007856  as  an  example. 

Beginning  at  the  decimal  point  and  pointing  off  into  periods  of  two 
figures  each,  we  have  .00'07'85'60.  Hence,  the  first  figure  of  the  root  must 
be  a  cipher.  To  obtain  the  rest  of  the  root  we  proceed  as  before. 

.00'07'85'60(.0280  +  ,  Answer. 
4 


40 
_8 
48 


385 


384 


560  |      1  60 

It  will  be  noticed  that  the  square  root  of  a  decimal  will  always 
be  a  decimal.  If  we  square  a  fraction,  we  will  get  a  smaller 
fraction  for  its  square,  (£)2  =  ^;  or  as  a  decimal,  .252  =  .0625. 
Therefore,  the  opposite  is  true;  that,  if  we  take  the  square  root 
of  a  number  entirely  a  decimal,  will  get  a  decimal,  but  it  will  be 
larger  than  the  one  of  which  it  is  the  square  root.  Notice  the 
example  just  given:  .0007856  is  less  than  its  square  root  .028. 

79.  Rules  for  Square  Root. — From  the  preceding  examples 
the  following  rules  may  be  deduced: 

1.  Beginning  at  the  decimal  point  separate  the  number  into 
periods  of  two  figures  each.     If  there  is  no  decimal  point  begin 
with  the  figure  farthest  to  the  right. 

2.  Find  the  greatest  whole  number  whose  square  is  contained 
in  the  first  or  left-hand  period.     Write  this  number  as  the  first 
figure  in  the  root;  subtract  the  square  of  this  number  from  the 
first  period,  and  annex  the  second  period  to  the  remainder. 

3.  Annex  a  cipher  to  the  part  of  the  root  already  found  and 
multiply  by  2;  this  gives  the  trial  divisor.     Divide  the  dividend 
by  the  trial  divisor  for  the  second  figure  of  the  root  and  add  this 
figure  to  the  trial  divisor  for  the  complete  divisor.     Multiply 
the  complete  divisor  by  the  second  figure  in  the  root  and  sub- 
tract this  result  from  the  dividend.     (If  this  result  is  larger  than 
the  dividend,  a  smaller  number  must  be  tried  for  the  second 
figure  of  the  root.) 

Bring  down  the  third  period  and  annex  it  to  the  last  remainder 
for  the  new  dividend. 

4.  Repeat  rule  3  until  the  last  period  is  used,  after  which, 
if  any  additional  decimal  places   are  required,   annex   cipher 
periods  and  continue  as  before.     If  the  last  period  in  the  decimal 
should  contain  but  one  figure,  annex  a  cipher  to  make  a  full 
period. 


SQUARE  ROOT 


89 


5.  If  at  any  time  the  trial  divisor  is  not  contained  in  the 
dividend,  place  a  cipher  in  the  root,  annex  a  cipher  to  the  trial 
divisor  and  bring  down  another  period. 

6.  To  locate  the  decimal  point,  remember  that  there  will  be 
as  many  figures  in  the  root  to  the  left  of  the  decimal  point  as 
there  were  periods  to  the  left  of  the  decimal  point  in  our  original 
number. 

80.  The  Law  of  Right  Triangles. — One  of  the  most  useful  laws 
of  geometry  is  that  relating  to  the  sides  of  a  right  angled  triangle. 
Fig.  28  shows  a  right  angled  triangle,  or  "right  triangle,"  so 


FIQ.  29. 


called  because  one  of  its  angles  (the  one  at  C)  is  a  right  angle,  or 
90°.  The  longest  side  (c)  is  called  the  hypotenuse.  "In  any 
right  triangle  the  square  of  the  hypotenuse  is  equal  to  the  sum  of 
the  squares  of  the  other  two  sides."  Written  as  a  formula  this 
would  read 


This  can  be  illustrated  by  drawing  squares  on  each  side,  as  in 
Fig.  29,  and  noting  that  the  area  of  the  square  on  the  hypotenuse 
is  equal  to  the  sum  of  the  areas  of  the  other  two. 

In  using  this  rule,  however,  we  do  not  care  anything  about  these 


90 


SHOP  ARITHMETIC 


areas  and  seldom  think  of  them  except  as  being  the  squares  of 
numbers.  It  is  used  to  find  one  side  of  such  a  triangle  when  the 
other  two  are  known. 

Examples  : 

1.  If  the  trolley  pole  in  Fig.  30  is  24  ft.  high,  and  the  guy  wire  is  anchored 
7  ft.  from  the  base  of  the  pole,  what  is  the  length  of  the  guy  wire? 

The  guy  wire  is  the  hypotenuse  of  a  right  triangle  whose  sides  are  24  ft. 
and  7  ft. 

c2  =  b'  +  a2 
c2  =  24-<  +  72 

=  576  +  49  =  625 
c  =  vX625  =  25  ft.,  Answer. 

2.  If  the  triangle  of  Fig.  31  is  a  right  triangle  having  the  hypotenuse 
c  =  13  in.  and  the  side  a  =  5  in.,  what  is  the  length  of  the  side  6? 


Hence, 


2  =  132-52 
=  169-25  =  144 


b  =\/144  =  12  in.,  Answer. 


This  property  of  right  triangles  is  also  useful  in  laying  out 
right  angles  on  a  large  scale  more  accurately  than  it  can  be  done 
with  a  square.  This  is  done  by  using  three  strings,  wires,  or 
chains  of  such  lengths  that  when  stretched  they  form  a  right 
triangle.  A  useful  set  of  numbers  that  will  give  this  are  3,  4,  and 
5,  since  32  +  42  =  52  (9  +  16-25). 

Any  three  other  numbers  having  the  same  ratios  as  3,  4? 
and  5  can  be  used  if  desired.  6,  8,  and  10;  9,  12,  and  15;  12,  16; 
and  20;  15,  20,  and  25;  any  of  these  sets  of  numbers  can  be  used. 


SQUARE  ROOT  91 

A  surveyor  will  often  use  lengths  of  15  ft.,  20  ft.,  and  25  ft.  on 
his  chain  to  lay  out  a  square  corner;  this  method  can  also  be  used 
in  aligning  engines,  shafting,  etc. 

81.  Dimensions  of  Squares  and  Circles.  —  Square  Root  must  be 
used  in  getting  the  dimensions  of  a  square  or  a  circle  to  have 
a  given  area.  If  the  area  of  a  square  is  given,  the  length  of  one 
side  can  be  obtained  by  extracting  the  square  root  of  the  area. 
If  we  wish  to  know  the  diameter  of  a  circle  which  shall  have  a 
certain  area,  we  can  find  it  by  the  following  process: 

The  area  is  .  7854  X  the  square  of  the  diameter  or,  briefly, 


If  we  divide  the  given  area  by  .7854,  we  will  get  the  area  of 
the  square  constructed  around  the  circle  (see  Fig.  19). 

One  side  of  this  square  is  the  same  as  the  diameter  of  the  circle 
and  is  equal  to  the  square  root  of  the  area  of  the  square. 

Then,  to  find  the  diameter  of  a  circle  to  have  a  given  area: 
Divide  the  given  area  by  .  7854  and  extract  the  square  root  of  the 
quotient. 

n-     P1" 

=  \77854~ 

82.  Dimensions  of  Rectangles.  —  Occasionally  one  encounters 
a  problem  in  which  he  wants  a  rectangle  of  a  certain  area  and 
knows  only  that  the  two  dimensions  must  be  in  some  ratio. 
It  may  be  that  a  factory  building  is  to  cover,  say  40,000  sq.  ft. 
of  ground  and  is  to  be  four  times  as  long  as  it  is  wide,  or  some 
problem  of  a  similar  nature.  Suppose  we  take  the  case  of  this 
factory  and  see  how  we  would  proceed  to  find  the  dimensions 
of  the  building. 

Example  : 

Wanted  a  factory  building  to  cover  40,000  sq.  ft.  of  ground. 
Ratio  of  length  to  breadth,  4:1.     Find  the  dimensions. 


I 

1 
1 

1 

1 
1 

40000  -s-4 
b 

Fio.  32. 

=  10000 
=  10000 
=  100 
=  4X100  =  400. 

92  SHOP  ARITHMETIC 

Explanation:  If  we  divide  the  total  area  by  4,  we  get  10,000  as  the  area  of 
a  square  having  the  breadth  6  on  each  side.  From  this  we  find  the  breadth 
or  width  6  to  be  the  square  root  of  10,000  or  100  ft.  If  the  length  is  four 
times  as  great  it  will  be  400  ft.  and  the  dimensions  of  the  building  will  be 
400  by  100. 

83.  Cube  Root.  —  The  Cube  Root  of  a  given  number  is  another 
number  which,  when  cubed,  produces  the  given  number.  In  other 
words,  the  cube  root  is  one  of  the  three  equal  factors  of  a  number. 
The  cube  root  of  8  is  2,  because  23  =  2x2x2  =  8;  also  the  cube 
root  of  27  is  3  (since  33  =  27)  and  the  cube  root  of  64  is  4  (since 
43  =  64). 

The  sign  of  cube  root  is  V  placed  over  the  number  of  which 
we  want  the  root.  Thus  we  would  write 

4  ^1000=  10 


If  we  consider  the  number  of  which  we  want  the  cube  root 
as  representing  the  volume  of  a  cubical  block,  then  the  cube 
root  of  the  number  will  represent  the  length  of  one  edge  of  the 
cube.  The  cube  root  of  1728  is  12  and  a  cube  containing  1728 
cu.  in.  will  measure  12  in.  on  each  edge. 

There  are  four  ways  of  getting  cube  roots:  (1)  by  actual 
calculation,  (2)  by  reference  to  a  table  of  cubes  or  cube  roots, 
(3)  by  the  use  of  logarithms,  and  (4)  by  the  use  of  some  calcu- 
lating device  like  the  slide  rule. 

The  use  of  a  table  is  the  simplest  way  of  finding  cube  roots, 
but  its  value  and  accuracy  is  limited  by  the  size  of  the  table. 
Tables  of  cubes  or  cube  roots  are  to  be  found  in  many  handbooks 
and  catalogues  and  should  be  used  whenever  they  give  the  desired 
root  with  sufficient  accuracy. 

Logarithms  give  us  an  easy  way  of  getting  cube  roots,  but 
here  also  a  table  is  necessary  and  the  accuracy  is  limited  by  the 
size  of  the  table  of  logarithms.  'The  use  of  logarithms  will  be 
explained  in  a  later  chapter.  The  ordinary  pocket  slide  rule 
will  give  the  first  three  figures  of  a  cube  root  and  for  many 
calculations  this  is  sufficiently  accurate.  The  method  of  actually 
calculating  cube  roots  is  very  complicated  and  is  used  so  seldom 
that  one  can  never  remember  it  when  he  needs  it.  Consequently, 
if  it  is  necessary  to  hunt  up  a  book  to  find  how  to  extract  the 
cube  root,  one  might  just  as  well  look  up  a  table  of  cube  roots 
or  a  logarithm  table,  either  of  which  will  give  the  root  much 
quicker.  The  next  chapter  contains  tables  of  cube  roots  and  a 
chapter  further  on  explains  the  use  of  logarithms. 


SQUARE  ROOT 


PROBLEMS 


93 


141.  Extract  the  square  roots  of 

(a)  64516  Answer  254 

(6)198.1369  Answer  14.076  + 

(c)   .571428  Answer  .7559  + 

Note. — These  answers  are  given  so  that  the  student  can  see  if  he 
understands  the  operations  of  square  root  before  proceeding 
further. 

142.  The  two  sides  of  a  right  triangle  are  36  and  48  ft. ;  what  is  the  length 
of  the  hypotenuse? 

143.  A  square  nut  for  a  2  in.  bolt  is  3£  in.  on  each  side.     What  is  the 
length  of  the  diagonal,  or  distance  across  the  corners? 

144.  A  steel  stack  75  ft.  high  is  to  be  supported  by  4  guy  wires  fastened 
to  a  ring  two-thirds  of  the  way  up  the  stack  and  having  the  other  ends 
anchored  at  a  distance  of  50  ft.  from  the  base  and  on  a  level  with  the  base. 
How  many  feet  of  wire  are  necessary,  allowing  20  ft.  extra  for  fastening  the 
ends? 

145.  What  would  be  the  diameter  of  a  circular  brass  plate  having  an 
area  of  100  sq.  in.? 


FIG.  33. 


146.  A  lineshaft  and  the  motor  which  drives  it  are  located  in  separate 
rooms  as  shown  in  Fig.  33.     Calculate  the  exact  distance  between  the 
centers  of  the  two  shafts. 

147.  I  want  to  cut  a  rectangular  sheet  of  drawing  paper  to  have  an  area 
of  235  sq.  in.  and  to  be  one  and  one-half  times  as  long  as  it  is  wide.     What 
would  be  the  dimensions  of  the  sheet? 

148.  A  6  in.  pipe  and  an  8  in.  pipe  both  discharge  into  a  single  header. 
Find  the  diameter  of  the  header  so  that  it  will  have  an  area  equal  to  that  of 
both  the  pipes. 

149.  What  would   be   the  diameter  of  a  1  Ib.  circular  cast  iron  weight 
J  in.  thick? 


94 


SHOP  ARITHMETIC 


150.  How  long  must  be  the  boom  in  Fig.  34  to  land  the  load  on  the  12  ft. 
pedestal,  allowing  4  ft.  clearance  at  the  end  for  ropes,  pulleys,  etc.? 


FIG.  34. 


CHAPTER  XII 
MATHEMATICAL  TABLES  (CIRCLES,  POWERS,  AND  ROOTS) 

84.  The  Value  of  Tables. — There  are  certain  calculations  that 
are  made  thousands  of  times  a  day  by  different  people  in  different 
parts  of  the  world.  For  example,  the  circumferences  of  circles 
of  different  diameters  are  being  calculated  every  day  by  hundreds 
and  thousands  of  men.  To  save  much  of  the  time  that  is  thus 
wasted  in  useless  repetition,  many  of  the  common  operations 
and  their  results  have  been  "tabulated,"  that  is,  arranged  in 
tables  in  the  same  way  as  are  our  multiplication  tables  in  arith- 
metics. These  tables  are  not  learned,  however,  as  were  the 
multiplication  tables,  but  are  consulted  each  time  that  we  have 
need  for  their  assistance. 

Just  what  tables  one  needs  most,  depends  on  his  occupation. 
The  machinist  has  use  for  tables  of  the  decimal  equivalents  of 
common  fractions,  tables  of  cutting  speeds,  tables  of  change 
gears  to  use  for  screw  cutting,  etc.  The  draftsman  would  use 
tables  of  strengths  and  weights  of  different  materials,  safe  loads 
for  bolts,  beams,  etc.,  tables  of  proportions  of  standard  machine 
parts  of  different  sizes,  etc.  The  engineer  uses  tables  of  the  prop- 
erties of  steam,  and  of  the  horse-power  of  engines,  boilers,  etc, 


MATHEMATICAL  TABLES  95 

There  are  certain  mathematical  tables  that  are  of  value  to 
nearly  everyone.  Among  these  are  the  tables  given  in  this 
chapter:  Tables  of  Circumferences  and  Areas  of  Circles;  Tables 
of  Squares,  Cubes,  Square  Roots,  and  Cube  Roots  of  Numbers. 

85.  Explanation  of  the  Tables. — The  first  table  is  to  save  the 
necessity  of  always  multiplying  the  diameter  by  3.1416  when  we 
want  the  circumference  of  a  circle,  or  of  squaring  the  diameter 
and  multiplying  by  .7854  when  the  area  of  a  circle  is  wanted. 

To  find  the  circumference  of  a  circle:  Find,  in  the  diameter 
column,  the  number  which  is  the  given  diameter;  directly 
across,  in  the  next  column  to  the  right,  will  be  found  the  corre- 
sponding circumference. 

Examples : 

Diameter    1J  Circumference      3.9270 

Diameter  27,  Circumference    84.823 

Diameter  90J,  Circumference  284.314 

To  find  the  area  of  a  circle:  Find,  in  the  diameter  column, 
the  number  which  is  the  given  diameter;  directly  across,  in  the 
second  column  to  the  right  (the  column  headed  "Area")  will  be 
found  the  area. 

Examples : 

Diameter  66,  Area  3421.2 

Diameter  17,  Area    226.98 

Diameter  f ,  Area        0.3067 

If  the  area  or  circumference  is  known  and  we  want  to  get  the 
diameter,  we  find  the  given  number  in  the  area  or  circumference 
column  and  read  the  diameter  in  the  corresponding  diameter 
column  to  the  left. 

Examples : 

Area  78.54,  Diameter  10 

Area  706.86,  Diameter  30 

Circumference  281.  Diameter  89  J 

The  second  table,  that  of  squares,  cubes,  square  roots,  and 
cube  roots,  is  especially  valuable  in  avoiding  the  tedious  process 
of  extracting  square  or  cube  roots.  The  table  is  read  the  same  as 
the  other  one.  Find  the  given  number  in  the  first  column;  on 
a  level  with  it,  in  the  other  columns,  will  be  found  the  corre- 
sponding powers  and  roots,  as  indicated  in  the  headings  at  the 
tops  of  the  columns. 


96  SHOP  ARITHMETIC 

Examples : 

^25  =     2.924 
\/260  -   16.1245 
2953=   25,672,375 
8652  =  748,225 

86.  Interpolation. — This  is  a  name  given  to  the  process  of 
finding  values  between  those  given  in  the  tables.  For  example, 
suppose  we  want  the  circumference  of  a  30 \  in.  circle.  The  table 
gives  30  and  30^  and,  since  30J  is  half  way  between  these,  its 
circumference  will  be  half  way  between  that  of  a  30  in.  and  a 
30J  in.  circle. 

Circumference  of  30 \  in.  circle  =  95. 8 19 
Circumference  of  30    in.  circle  =  94 . 248 


The  Difference  =    1.571 

Then  the  circumference  of  the  30 \  in.  circle  is  just  half  this 
difference  more  than  that  of  the  30-in.  circle, 

94.248  +  ^  of  1.571  =  95.034 

A 

This  method  enables  us  to  increase  greatly  the  value  of  tables. 
For  most  purposes  the  interpolation  can  be  done  quickly,  and 
while  it  requires  some  calculating,  is  much  shorter  than  the 
complete  calculation  would  be.  This  is  especially  true  in  finding 
square  or  cube  roots. 

Example : 

Find  from  the  table  the  cube  root  of  736.4 
3/737  =  9.0328 
=  9.0287 


Difference =41 

.4  X  the  difference  =  .4X41  =  16.4 
9.0287 
16 
^736>4=  9.0303,  Answer. 

Explanation:  The  root  of  736.4  will  be  between  that  of  736  and  that  of 
737,  and  will  be  .4  of  the  difference  greater  than  that  of  736.  In  making 
this  correction  for  the  .4,  we  forget,  for  the  minute,  that  the  difference  is  a 
decimal  and  write  it  as  41  merely  to  save  time.  We  then  multiply  it  by  .4, 
and,  dropping  the  decimal  part,  add  the  16  to  the  90287.  This  gives  9.0303 
as  the  cube  root  of  736.4 

Hence, 


MATHEMATICAL  TABLES  97 

87.  Roots  of  Numbers  Greater  than  1000.— For  getting  the 
cube  roots  of  numbers  greater  than  1000,  the  easiest  and  most 
accurate  way  is  to  look  in  the  third  column  headed  "cubes"  for 
a  number  as  near  as  possible  to  our  given  number.  Now,  we 
know  that  the  numbers  in  the  first  column  are  the  cube  roots  of 
these  numbers  in  the  third  column.  If  we  can  find  our  number 
in  the  third  column,  there  is  nothing  further  to  do  because  its 
cube  root  will  be  directly  opposite  it  in  the  first  column. 

Examples : 

3/1728       =  12 
3/6967871  =  191 
3/166375   =55 

Likewise,  the  numbers  in  the  first  column  are  the  square  roots  of 
the  numbers  in  the  second  column.  But  suppose  we  want  the 
cube  root  of  a  number  which  is  not  found  in  the  third  column, 
but  lies  somewhere  between  two  consecutive  numbers  in  that 
column.  In  this  case  we  pursue  the  method  shown  in  the 
following  example: 

Example : 

Find  3/621723 

In  the  column  headed  "  Cube"  find  two  consecutive  numbers,  one  larger  and 
one  smaller  than  621723.  These  numbers  are 

636056  whose  cube  root  is  86 
and      614125  whose  cube  root  is  85 

Hence,  the  cube  root  of  621723  is  more  than  85  and  less  than  86;  that  is,  it 
is  85  and  a  decimal,  or  85  +  . 

The  decimal  part  is  found  as  follows:  Subtract  the  lesser  of  the  two  num- 
bers found  in  the  table  from  the  greater  and  call  the  result  the  First 
Difference. 

636056-614125  =  21931,  First  difference. 

Then  subtract  the  smaller  of  the  two  numbers  in  the  table  from  the  given 
number  and  call  the  result  the  Second  Difference. 

621723-614125  =  7598,  Second  Difference 

Now  the  first  difference,  21931,  is  the  amount  that  the  number  increases 
when  its  cube  root  changes  from  85  to  86.  Our  given  number  is  only  7598 
more  than  the  cube  of  85,  so  its  cube  root  will  be  approximately  85  zWVr- 
We  do  not  want  a  fraction  like  this,  so  we  reduce  it  to  a  decimal  as  follows: 
Divide  the  second  difference  by  the  first  difference  and  annex  the  quotient 
to  85.  This  will  give  us  the  cube  root  of  our  number,  approximately. 

Second  Difference  _   7598  _   ., .  „ 
First      Difference  "  2l93l " 

This  is  the  decimal  part  of  the  root  sought  and  the  whole  root  is  85.346  + . 
Hence  3/621723  =  85.346  +  . 

This  method  is  not  exact  and  the  third  decimal  place  will  usually  be 
slightly  off,  so  it  is  best  to  drop  the  third  decimal  if  less  than  5,  or  raise  it  to 
10,  if  more  than  5.  In  this  case  we  will  call  the  root  85.35. 


98  SHOP  ARITHMETIC 

88.  Cube  Roots  of  Decimals. — In  getting  the  cube  root  of 
either  a  number  entirely  decimal,  or  a  mixed  decimal  number, 
it  is  best  to  move  the  decimal  point  a  number  of  periods,  that  is, 
3,  6,  9,  or  12  decimal  places,  sufficient  to  make  a  whole  number 
out  of  the  decimal.  After  finding  the  cube  root,  shift  the  decimal 
point  in  the  root  back  to  the  left  as  many  places  as  the  number  of 
periods  that  we  moved  the  decimal  point  in  our  original  number. 
For  example,  suppose  that  we  had  .621723  of  which  to  find  the 
cube  root.  Moving  the  decimal  point  two  periods  (of  three 
places  each)  to  the  right  gives  us  621723,  of  which  we  just  found 
the  cube  root  to  be  85.35.  We  moved  the  decimal  point  of  our 
original  number  two  periods  to  the  right,  so  we  must  move  the 
decimal  point  back  two  places  to  the  left  in  the  root;  we  then 
have  ^.621723  =  .8535.  The  following  illustrations  will  show 
the  principle: 


4/621723         =85.35 
V62  1.723       =  8.535 
V.  621723  .8535 

</•  00062  1723  =      .08535 

Notice  that  there  is  no  such  similarity  between  the  cube  roots  of 
numbers  if  we  move  the  decimal  point  any  other  number  of 
places  than  a  multiple  of  three. 

VQ=   1.817+          V60=  3.915  ^600=   8.434 

But  ^6000  =  18.17        ^60,000  =  39.  15      ^600,000  =  84.34 

Care  should  be  taken,  therefore,  that,  if  necessary  to  move  the 
decimal  point  in  finding  a  cube  root,  it  should  be  moved  an  exact 
multiple  of  3  places.  If  we  have  a  decimal  such  as  .07462,  it  is 
necessary  to  attach  a  cipher  at  the  right,  making  the  decimal 
.074620,  so  we  can  shift  the  decimal  point  2  periods  or  six  places. 
We  can  now  find  v  74620  as  follows: 

433  =  79507  74620 

423  =  74088  74088 

5419  First  Difference.          532  Second  Difference. 
532 


Hence  ^74620  =  42.01 
and  V.  074620=      .4201 


MATHEMATICAL  TABLES  99 

89.  Square  Root  by  the  Table. — The  same  methods  as  have 
been  explained  for  cube  root  can  be  applied  to  finding  square 
roots  by  the  use  of  the  table.  The  only  difference  is  in  the  case 
of  decimals,  in  which  case  the  decimal  point  is  shifted  by  mul- 
tiples of  two  places  in  the  number;  then,  after  we  have  the  root,  we 
shift  the  decimal  point  back  one  place  for  each  period  of  two 
places  that  we  moved  the  decimal  point  in  our  original  number. 


PROBLEMS 

The  tables  are  to  be  used  wherever  possible  in  working  these  problems. 

151.  What  would  be  the  length  of  a  steel  sheet  from  which  to  make  a 
28  in.  circular  drum,  allowing  1^  in.  extra  for  lapping  and  riveting  the  ends? 
The  circumference  of  the  drum  is  measured  in  the  direction  of  the  length  of 
the  sheet. 

152.  The  small  sprocket  of  a  bicycle  contains  8  teeth,  the  large  sprocket 
24  teeth.     The  rear  wheel  is  30  in.  in  diameter.     Find  the  distance  travelled 
over  by  the  bicycle  for  one  revolution  of  the  pedals. 

153.  The  diameter  of  a  If  in.  bolt  at  the  bottom  of  the  threads  is  1.16  in. 
What  is  the  sectional  area  at  the  bottom  of  the  threads? 

154.  A  No.  000  copper  trolley  wire  has  a  diameter  of  0.425  in.     What 
would  be  the  cost  of  1  mile  of  the  wire  at  33  cents  a  pound? 


155.  A  circular  piece  of  boiler  plate  (Fig.  35)  £  in.  thick  and  60  in.  diam- 
eter has  an  elliptical  man  hole  in  it  14  in.  by  10  in.     Find  weight  of  plate. 

Note. — Area  of  an  ellipse  =.7854  X«  X&,  where  a  and  6  are  the  long  and 
short  diameters  of  the  ellipse. 

156.  The  paint  shop  wants  a  cubical  dip  tank  built  to  hold,  when  full, 
400  gallons  of  varnish.     What  will  be  the  dimensions  of  the  tank?     (There 
are  231  cu.  in.  in  a  gallon.) 

157.  How  many  square  feet  of  galvanized  iron  will  be  needed  to  line  the 
tank  of  problem  156  on  four  sides  and  the  bottom,  allowing  10%  extra  for 
the  joints? 


100 


SHOP  ARITHMETIC 


158.  A  high  carbon  steel  contains  the  following  items  in  the  percentages 
given:  Carbon,  .60%;  silicon,  .10%;  manganese,  .40%;  phosphorus,  .035%; 
sulphur,  .025%.     The  rest  is  pure  iron.     Calculate  the  weights  of  carbon, 
silicon,  manganese,  phosphorus,  sulphur,  and  iron  in  one  ton  (2000  Ib.)  of 
the  steel. 

159.  I  want  to  get  a  cast  iron  block  18  in.  long  and  of  square  cross-section 
so  that  it  will  weigh  200  Ib.     How  many  cubic  inches  of  metal  must  be  in 
the  block  and  what  will  be  its  dimensions? 

160.  Fig.  36  shows  a  steam  hammer  having  an  8000  Ib.  ram.     If  we 
assume  that  this  ram  is  a  rectangular  block  of  steel,  four  times  as  high  and 
twice  as  wide  as  it  is  thick,  what  will  be  the  dimensions  of  the  ram? 


.  36. 


MATHEMATICAL  TABLES 
CIRCUMFERENCES  AND  AREAS  OF  CIRCLES 


101 


Diameter 

Circumference 

Area 

Diameter 

Circumference 

Area 

i 

.3927 

0.0123 

191 

61.261 

298.65 

I 

.7854 

0.0491 

20 

62.832 

314.16 

* 

1.1781 

0.1104 

201                  64.403 

330.06 

1 

1.5708 

0.1963 

21 

65.973 

346.36 

i 

1.9635 

0.3067 

211 

67.544 

363.05 

i 

2.3562 

0.4417 

22 

69.115 

380.13 

i 

2.7489 

0.6013 

221 

70.686 

397.61 

i 

3.1416 

0.7854 

23 

72.257 

415.48 

11 

3.5343 

0.9940 

231 

73.827 

433.74 

ii 

3.9270 

1.227 

24 

75.398 

452.39 

U 

4.3197 

1.484 

241 

76.969 

471.44 

H 

4.7124 

1.767 

25 

78.540 

490.87 

li 

5.1051 

2.073 

251 

80.111 

510.71 

U 

5.4978 

2.405 

26 

81.681 

530.93 

ii 

5.8905 

2.761 

261 

83.252 

551.55 

2 

6.2832 

3.141 

27 

84.823 

572.56 

21 

7.0686 

3.976 

271 

86.394 

593.96 

21 

7.8540 

4.908 

28 

87.965 

615.75 

2| 

8.6394 

5.939 

281 

89.535 

637.04 

3 

9.4248 

7.068 

29 

91.106 

660.52 

3i 

10.210 

8.295 

291 

92.677 

683.49 

31 

10.996 

9.621 

30 

94.248 

706.86 

3! 

11.781 

11.044 

301 

95.819 

730.62 

4 

12.566 

12.566 

31 

97.389 

754.77 

41 

14.137 

15.904 

311 

08.960 

779.31 

5 

15.708 

19.635 

32 

100.531 

804.25 

51 

17.279 

23.758 

321 

102.102 

829.58 

6 

18.850 

28.274 

33 

103.673 

855.30 

6i 

20.420 

33.183 

331 

105.243 

881.41 

7 

21.991 

38.485 

34 

106.814 

907.92 

71 

23.562 

44.179 

341 

108.385 

934.82 

8 

25.133 

50.265 

35 

109.956 

962.11 

81 

26.704 

56.745 

351 

111.527 

989.80 

9 

28.274 

63.617 

36 

113.097 

1017.9 

91 

29.845 

70.882 

361 

114.668 

1046.3 

10 

31.416 

78.540 

37 

116.239 

1075.2 

101 

32.987 

86.590 

371 

117.810 

1104.5 

11 

34.558 

95.033 

38 

119.381 

1134.1 

H* 

36.128 

103.87 

381 

120.951 

1164.2 

12 

37.699 

113.10 

39 

122.522 

1194.6 

121 

39.270 

122.72 

391 

124.093 

1225.4 

13 

40.841 

132.73 

40 

125.664 

1256.6 

131 

42.414 

143.14 

401 

127.235 

1288.2 

14 

43.982 

153.94 

41 

128.805 

1320.3 

141 

45.553 

165.13 

411 

130.376 

1352.7 

15 

47.124 

176.71 

42 

131.947 

1385.4 

151 

48.695 

188.60 

421 

133.518 

1418.6 

16 

50.265 

201.06 

43 

135.088 

1452.2 

161 

51.836 

213.82 

431 

136.659 

.      1486.2 

17 

53.407 

226.98 

44 

138.230 

1520.5 

171 

54.978 

240.53 

441 

139.801 

1555.3 

18 

56.549 

254.47 

45 

141.372 

1590.4 

181 

58.119 

268.80 

451 

142.942 

1626.0 

19 

59.690       • 

283.53 

46 

144.513 

1661.9 

102  SHOP  ARITHMETIC 

CIRCUMFERENCES  AND  AREAS  OF  CIRCLES.— Continue* 


Diameter 

Circumference 

Area 

Diameter 

Circumference 

Area 

46J 

146.084 

1698.2        731 

230.907 

4242.9 

47 

147.655 

1734.9 

74 

232.478 

4300.8 

47J 

149.226 

1772.1 

74} 

234.049 

4359.2 

48 

150.796 

1809.6 

75 

235.619 

4417.9 

481 

152.367 

1847.5 

75} 

237.190 

4477.0 

49 

153.938 

1885.7 

76 

238.761 

4536.5 

491 

155.509 

1924.4 

76} 

240.332 

4596.3 

50 

157.080 

1963.5 

77 

241.903 

4656.6 

501 

158.650 

2003.0 

77} 

243.473 

4717.3 

51 

160.221 

2042.8 

78 

245.044 

4778.4 

511 

161.792 

2083  .  1 

78} 

246.615 

4839.8 

52 

163.363 

2123.7 

79 

248.186 

4901.7 

521 

164.934      2164.8 

79} 

249.757 

4963.9 

53 

166.504       2206.2 

80 

251.327 

5026.5 

531 

168.075 

2248.0 

80} 

252.898 

5089.6 

54 

169.646 

2290.2 

81 

254.469 

5153.0 

541 

171.217 

2332.8 

81} 

256.040 

5216.8 

55 

172.788 

2375.8 

82 

257.611 

5281.0 

551 

174.358 

2419.2 

82}       259.181 

5345.6 

56 

175.929 

2463.0        83        260.752 

5410.6 

561 

177.500 

2507.2 

83} 

262.323 

5476.0 

57 

179.071 

2551.8 

84 

263.894 

5541.8 

571 

180.642 

2596.7 

84} 

265.465 

5607.9 

58 

182.212 

2642  .  1 

85 

267.035 

5674.5 

581 

183.783 

2687.8 

85} 

268.606 

5741.5 

59 

185.354 

2734.0 

86 

270.177 

5808.8 

591 

186.925 

2780.5 

86} 

271.748 

5876.5 

60 

188.496 

2827.4 

87 

273.319 

5944.7 

601 

190.066 

2874.8 

87} 

274.889 

6013.2 

61 

191.637 

2922.5 

88 

276.460 

6082.1 

611 

193.208 

2970.6 

88} 

278.031 

6151.4 

62 

194.779 

3019.1 

89 

279.602 

6221.1 

621 

196.350 

3068.0 

89} 

281.173 

6291  .2 

63 

197.920 

3117.2 

90 

282.743 

6361.7 

631 

199.491 

3166.9 

90} 

284.314 

6432.6 

64 

201.062 

3217.0 

91 

285.885 

6503.9 

641 

202.633 

3267.5 

91} 

287.456 

6575.5 

65 

204.204 

3318.3 

92 

289.027 

6647.6 

651 

205.774 

3369.6 

92} 

290.597 

6720.1 

66 

207.345 

3421.2 

93 

292.168 

6792.9 

661 

208.916 

3473.2 

93} 

293.739 

6866.1 

67 

210.487 

3525.7 

94 

295.310 

6939.8 

671 

212.058 

3578.5 

94} 

296.881 

7013.8 

68 

213.628 

3631.7 

95 

298.451 

7088.2 

681 

215.199 

3685.3        951 

300.022 

7163.0 

69 

216.770 

3739.3        96 

301.593 

7238.2 

691 

218.311 

3793.7        961 

303.164 

7313.8 

70 

219.911 

3848.5        97 

304.734 

7389.8 

701 

221.  4S2 

3903.6        971 

306.305 

7466.2 

71 

223.053 

3959.2        98 

307.876 

7543.0 

711 

224.624 

4015.2        981 

309.447 

7620.1 

72 

226.195 

4071.5        99 

311.018 

7697.7 

721 

227.765 

4128.2        991 

312.588 

7775.6 

73 

229.336 

4185.4 

100 

'  314.159 

7854.0 

MATHEMATICAL  TABLES 


103 


SQUARES,   CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 

NUMBERS 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

1 

1 

1 

1. 

1. 

46 

2116 

97336 

6.7823 

3.5830 

2 

4 

8 

1.4142 

1.2599 

47 

2209 

103823 

6.8557 

3.6088 

3 

9 

27 

1.7321 

1.4422 

48 

2304 

110592 

6.9282 

3.6342 

4 

16 

64 

2. 

1.5874 

49 

2401 

117649 

7. 

3.6593 

5 

25 

125 

2.2361 

1.7100 

50 

2500 

125000 

7.0711 

3.6840 

6 

36 

216 

2.4495 

1.8171 

51 

2601 

132651 

7.1414 

3.7084 

7 

49 

343 

2.6458 

1.9129 

52 

2704 

140608 

7.2111 

3.7325 

8 

64 

512 

2.8284 

2. 

63 

2809 

148877 

7.2801 

3.7563 

9 

81 

729 

3. 

2.0801 

54 

2916 

157464 

7.3485 

3.7798 

10 

100 

1000 

3.1623 

2.1544 

55 

3025 

166375 

7.4162 

3.8030 

11 

121 

1331 

3.3166 

2.2240 

56 

3136 

175616 

7.4833 

3.8259 

12 

114 

1728 

3.4641 

2.2894 

57 

3249 

185193 

7.5498   3.8485 

13 

169 

2197 

3.6056 

2.3513 

58 

3364 

195112 

7.6158   3.8709 

14 

196 

2744 

3.7417 

2.4101 

59 

3481 

205379 

7.6811   3.8930 

15 

225 

3375 

3.8730 

2.4662 

60 

3600 

216000 

7.7460 

3.9149 

16 

256 

4096 

4. 

2.5198 

61 

3721 

226981 

7.8102 

3.9365 

17 

289 

4913 

4.1231 

2.5713 

62 

3844 

238328 

7.8740   3.9579 

18 

324 

5832 

4.2426 

2.6207 

63 

3969 

250047 

7.9373 

3.9791 

I'.i 

361 

6859 

4.3589 

2.6684 

64 

4096 

262144 

8. 

4. 

20 

400 

8000 

4^4721 

2.7144 

65 

4225 

274625 

8.0623 

4.0207 

21 

441 

9261 

4.5826 

2.7589 

66 

4356 

287496 

8.1240 

4.0412 

22 

484 

10648 

4.6904 

2.8020 

67 

4489 

300763 

8.1854 

4.0615 

23 

529 

12167 

4.7958 

2.8439 

68 

4624 

314432 

8.2462 

4.0817 

21 

576 

13824 

4.8990 

2.8845 

69 

4761 

328509 

8.3066 

4.1016 

25 

625 

15625 

5. 

2.9240 

70 

4900 

343000 

8.8666 

4.1213 

26 

676 

17576 

5.0990 

2.9625 

71 

5041 

357911 

8.4261 

4.1408 

27    729     19683   5.1962 

Q 

72 

5184 

373248 

8.4853 

4.1602 

28    781     21952   5.2915 

3.0366 

73 

5329 

389017 

8.5440 

4.1793 

29    84  li     24389   5.3852 

3.0723 

74 

5476 

405224 

8.6023 

4.1983 

30    900 

27000 

5.4772 

3.1072 

75 

5625 

421875 

8.6603 

4.2172 

31    961 

29791 

5.5678 

3.1414 

76 

6776 

438976 

8.7178 

4.2358 

32    1024 

32768 

5.6569 

3.1748 

77 

5929 

456533 

8.7750 

4.2543 

33    1089 

35937 

5.7446 

3.2075 

78 

6084 

474552 

8.8318 

4.2727 

34    1156 

39304 

5.8310 

3.2396 

79 

6241 

493039 

8.8882 

4.2908 

35   1225 

42875 

5.9161 

3.2711 

80 

6400 

512000 

8.9443 

4.3089 

36   1296 

46656 

6. 

3.3019 

81 

6561 

631441 

9. 

4.3267 

37 

1369 

50653 

6.0828 

3.3322 

82 

6724 

551368 

9.0554 

4.3445 

38 

1444 

54872 

6.1614 

3.3620 

83 

6889 

571787 

9.1104 

4.3621 

39 

1521 

59319 

6.24.50 

3.3912 

84 

7056 

592704 

9.1652 

4.3795 

40 

1600 

64000 

6.3246 

3.4200 

85 

7225 

614125 

9.2195 

4.3968 

41 

1681 

68921 

6.4031 

3.4482 

86 

7396 

636056 

9.2736 

4.4140 

42 

1761     74088 

6.4807   3.4760 

87 

7569    658503 

9.3274 

4.4310 

43 

1849     79507   6.5574   3.5034J   88 

7744 

681472 

9.3808 

4.4480 

44 

1936     85184!  6.6332   3.5303] 

89 

7921 

704969 

9.4340 

4.4647 

45 

2025 

91125   6.7082 

3.55691  90 

8100 

729000 

9.4868 

4.4814 

104 


SHOP  ARITHMETIC 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
NUMBERS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

91 

8281 

753571 

9.5394 

4.4979 

136 

18496 

2515456 

11.6619 

5.1426 

92 

8464 

778688 

9.5917 

4.5144 

137 

18769 

2571353 

11.7047 

5.1551 

93 

8649 

804357 

9.6437 

4.5307 

138 

19044 

2628072 

11.7473 

5.1676 

91 

8836 

830584 

9.6954 

4.5468 

139 

19321 

2685619 

11.7898 

5.1801 

95 

9025 

857375 

9.7468 

4.5629 

140 

19600 

2744000 

11.8322 

5.1925 

96 

9216 

884736 

9.7980 

4.5789 

141 

19881 

2803221 

11.8743 

5.2048 

97 

9409 

912673 

9.8489 

4.5947 

142 

20164 

2863288 

11.9164 

5.2171 

98 

9604 

941192 

9.8995 

4.6104 

143 

20449 

2924207 

11.9583 

5.2293 

99 

9801 

970299 

9.9499 

4.6261 

144 

20736 

2985984 

12. 

5.2415 

100 

10000 

1000000 

10. 

4.6416 

145 

21025 

3048625 

12.0416 

5.2536 

101 

10201 

1030301 

10.0499 

4.6570 

146 

21316 

3112136 

12.0830 

5.2656 

102   10404 

1061208 

10.0995 

4.6723 

147 

21609 

3176523 

12.1244 

5.2776 

103   10609 

1092727 

10.1489 

4.6875 

148 

21904 

3241792 

12.1655 

5.2896 

10J   10816 

1124864 

10.1980 

4.7027 

149 

22201 

3307949 

12.2066 

5.3015 

105 

11025 

1157625 

10.2470 

4.7177 

150 

22500 

3375000 

12.2474 

5.3133 

106 

11236 

1191016 

10.2956 

4.7326 

151 

22801 

3442951 

12.2882 

5.3251 

107   11449 

1225043 

10.3441 

4.7475 

152 

23104 

3511808 

12.3288 

5.3368 

108   11664 

1259712 

10.3923 

4.7622 

153 

23409 

3581577 

12.3693 

5.3485 

109 

11881 

1295029 

10.4403 

4.7769 

154 

23716 

3652264 

12.4097 

5.3601 

110 

12100 

1331000 

10.4881 

4.7914 

155 

24025 

3723875 

12.4499 

5.3717 

111 

•  12.321 

1367631 

10.5357 

4.8059 

156 

24336 

3796416 

12.4900 

5.3832 

112 

12544 

1404928 

10.5830 

4.8203 

157 

24649 

3869893 

12.5300 

5.3947 

113 

12769 

1442897 

10.6301 

4.8346 

158 

24964 

3944312 

12  .  5698 

5.4061 

114 

12996 

1481544 

10.6771 

4.8488 

159 

25281 

4019679 

12.6095 

5.4175 

115 

13225 

1520875 

10.7238 

4.8629 

160 

25600 

4096000 

12.6491 

5.4288 

116 

13456 

1560896 

10.7703 

4.8770 

161 

25921 

4173281 

12.6886 

5.4401 

117 

13689 

1601613 

10.8167 

4.8910 

162 

26244 

4251528 

12.7279 

5.4514 

118   13924 

1643032 

10.8628 

4.9049 

163 

26569 

4330747 

12.7671 

5.4626 

119   14161 

1685159 

10.9087 

4.9187 

164 

26896 

4410944 

12.8062 

5.4737 

120 

•14400 

1728000 

10.9545 

4.9324 

165 

27225 

4492125 

12.8452 

5.4848 

121 

14641 

1771561 

11. 

4.9461 

166 

27556 

4574296 

12.8841 

5.4959 

122 

14884 

1815848 

11.0454 

4.9597 

167 

27889 

4657463 

12.9228 

5.5069 

123 

15129 

1860867 

11.0905 

4.9732 

168 

28224 

4741632 

12.9615 

5.5178 

124 

15376 

1906624 

11.1355 

4.9866 

169 

28561 

4826809 

13. 

5.5288 

125 

15625 

1953125 

11.1803 

5. 

170 

28900 

4913000 

13.0384 

5.5397 

126 

15876 

2000376 

11.2250 

5.0133 

171 

29241 

5000211 

13.0767 

5.5505 

127 

16129 

2048383 

11.2694 

5.0265 

172 

29584 

5088448 

13.1149 

5.5613 

128]  16384 

2097152 

11.3137 

5.0397 

173 

29929 

5177717 

13.1529 

5.5721 

129 

16641 

2146689 

11.3578 

5.0528 

174 

30276 

5268024 

13.1909 

5.5828 

130 

16900 

2197000 

11.4018 

5.0658 

175 

30625 

5359375 

13.2288 

5.5934 

131 

17161 

2248091 

11.4455 

5.0788 

176 

30976 

5451776 

13.2665 

5.6041 

132 

17424 

2299968 

11.4891 

5.0916 

177 

31329 

5545233 

13.3041 

5.6147 

133 

17689 

2352637 

11.5326 

5.1045 

178 

31684 

5639752 

13.3417 

5.6252 

134 

17956 

2406104 

11.5758 

5.1172 

179 

32041 

5735339 

13.3791 

5.6357 

135 

18225 

2460375 

11.6190 

5.1299 

180 

32400 

5S32000 

13.4164 

5.6462 

MATHEMATICAL  TABLES 


105 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
NUMBERS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

181 

32761 

5929741 

13.4536 

5.6567 

226 

51076 

11543176 

15.0333 

6.0912 

182 

33124 

6028568 

13.4907 

5.6671 

227 

51529 

11697083 

15.0665 

6.1002 

183 

33489 

6128487 

13.5277 

5.6774 

228 

51984 

11852352  15.0997 

6.1091 

184 

33856 

6229504 

13.5647 

5.6877 

229 

52441 

12008989  15.1327 

6.1180 

185 

34225 

6331625 

13.6015 

5.6980 

230 

52900 

12167000 

15.1658 

6.1269 

186 

34596 

6434856 

13.6382 

5.7083 

231 

53361 

12326391 

15.1987 

6.1358 

187 

34969 

6539203 

13.6748 

5.7185 

232 

53824 

12487168  15.2315 

6.1446 

188 

35344 

6644672 

13.7113 

5.7287 

233 

54289 

12649337  15.2643 

6.1534 

189 

35721 

6751269 

13.7477 

5.7388 

234 

54756 

12812904 

15.2971 

6.1622 

190 

36100 

6859000 

13.7840 

5.7489 

235 

55225 

12977875 

15.3297 

6.1710 

191 

36481 

6967871 

13.8203 

5.7590 

236 

55696 

13144256 

15.3623 

6.1797 

192 

36864 

7077888 

13.8564 

5.7690 

237 

56169 

13312053 

15.3948 

6.1885 

193 

37249 

7189057 

13.8924 

5.7790 

238 

56644 

13481272")  15.4272 

6.1972 

194 

37636 

7301384 

13.8284 

5.7890 

239 

57121 

13651919  15.4596 

6.2058 

195 

38025 

7414875 

13.9642 

5.7989 

240 

57600 

13824000 

15.4919 

6.2145 

196 

38416 

7529536 

14. 

5.8088 

241 

58081 

13997521 

15.5242 

6.2231 

197 

38809 

7645373 

14.0357 

5.8186 

242 

58564 

14172488 

15.5563 

6.2317 

198 

39204   7762392 

14.0712 

5.8285 

243 

59049 

14348907 

15.5885 

6.2403 

l-.r.i 

39601   7880599   14.1067 

5.8383 

244 

59536 

14526784 

15.6205 

6.2488 

200 

40000 

8000000 

14.1421 

5.8480 

245 

60025 

14706125 

15.6525 

6.2573 

201 

40401 

8120601 

14.1774 

5.8578 

246 

60516 

14886936 

15.6844 

6.2658 

202 

40804 

8242408 

14.2127 

5.8675 

247 

61009 

•  15069223 

15.7162 

6.2743 

203 

41209 

8365427 

14.2478 

5.8771 

248 

61504 

15252992 

15.7480 

6.2828 

204 

41616 

8489664 

14.2829 

5.8868 

249 

62001 

15438249 

15.7797 

6.2912 

205 

42025 

8615125 

14.3178 

5.8964 

250 

62500 

15625000 

15.8114 

6.2996 

206 

42436 

8741816 

14.3527 

5.9059 

251 

63001 

15813251 

15.8430 

6.3080 

207 

42849 

8869743 

14.3875 

5.9155 

252 

63504 

16003008 

15.8745 

6.3164 

208 

43264 

8998912 

14.4222 

5.9250 

253 

64009 

16194277 

15.9060 

6.3247 

209 

43681 

9129329 

14.4568 

5.9345 

254 

64516 

16387064 

15.9374 

6.3330 

210 

44100 

9261000 

14.4914 

5.9439 

255 

65025 

16581375 

15.9687 

6.3413 

211 

44521 

9393931 

14.5258 

5.9533 

256 

65536 

16777216 

16. 

6.3496 

212 

44944 

9528128 

14.5602 

5.9627 

257 

66049 

16974593 

16.0312 

6.3579 

213 

45369 

9663597 

14.5945 

5.9721 

258 

66564 

17173512 

16.0624   6.3661 

214 

45796 

9800344 

14.6287 

5.9814 

259 

67081 

17373979 

16.0935 

6.3743 

215 

46225 

9938375 

14.6629 

5.9907 

260 

67600 

17576000 

16.1245 

6.3825 

216 

46656 

10077696 

14.6969 

6. 

261 

68121 

17779581 

16.1555 

6.3907 

L'17 

47089 

10218313 

14.7309 

6.0092 

262 

68644 

17984728 

16.1864 

6.3988 

218 

47524 

10360232 

14.7648 

6.0185 

263 

69169 

18191447 

16.2173 

6.4070 

219 

47961 

10503459 

14.7986 

6.0277 

264 

69696 

18399744 

16.2481 

8.4151 

220 

48400 

10648000 

14.8324 

6.0368 

265 

70225 

18609625 

16.2788 

6.4232 

221 

48841 

10793861 

14.8661 

6.0459 

266 

70756 

18821096 

16.3095 

6.4312 

222 

49284 

10941048  14.8997 

6.0550 

267 

71289 

19034163 

16.3401 

6.4393 

223 

49729   11089567  14.9332 

6.0641 

268 

71824 

19248832 

16.3707 

6.4473 

224 

50176   11239424  14.9666 

6.0732 

269 

72361 

19465109 

16.4012 

6.4553 

225 

50625   11390625  15. 

6.0822 

270 

72900   19683000 

16.4317 

6.4633 

106 


SHOP  ARITHMETIC 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
NUMBERS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

271 

73441 

19902511 

16.4621 

6.4713 

316 

99856 

31554496 

17.7764 

6.8113 

272 

73984 

20123648 

16.4924 

6.4792 

317 

100489 

31855013 

17.8045 

6.8185 

273 

74529 

20346417 

16.5227 

6.4872 

318 

101124 

32157432  17.8326 

6.8256 

274 

75076 

20570824 

16.5529 

6.4951 

319 

101761 

32461  759  j  17.8606 

6.8328 

275 

75625 

20796875 

16.5831 

6.5030 

320 

102400 

32768000'  17.8885 

6.8399 

276   76176 

21024576 

16.6132 

6.5108 

321 

103041 

33076161 

17.9165 

6.8470 

277   76729 

21253933 

16.6433 

6.5187 

322 

103681 

33386248  17.9444 

6.8541 

278   77284 

21484952 

16.6733 

6.5265 

323 

104329 

33698267 

17.9722 

6.8612 

279   77841 

21717639 

16.7033 

6.5343 

324 

104976 

34012224 

18. 

6.8683 

280   78400 

21952000 

16.7332 

6.5421 

325 

105625 

34328125 

18.0278 

6.8753 

281 

78961 

22188041 

16.7631 

6.5499 

326 

106276 

34645976 

18.0555 

6.8824 

282   79524 

22425768  16.7929 

6.5577 

327 

106929 

34965783  18.0831 

6.8894 

283   80089 

22665187  16.8226 

6.5654 

328 

107584 

35287552  18.1108 

6.8964 

284   80656 

22906304  16.8523 

6.5731 

329 

108241 

35611289 

18.1384 

6.9034 

285   81225 

231491251  16.8819 

6.5808 

330 

108900 

35937000 

18.1659 

6.9104 

286 

81796 

23393656 

16.9115 

6.5885 

331 

109561 

36264691 

18.1934 

6.9174 

287   82369 

23639903  16.9411 

6.5962 

332 

110224   36594368 

18.2209 

6.9244 

288   82944 

23887872  16.9706 

6.6039 

333 

110889   36926037 

18.2483 

6.9313 

289   83521 

24137569  17. 

6.6115 

334 

111556 

37259704 

18.2757 

6.9382 

290   84100 

24389000 

17.0294 

6.6191 

335 

112225 

37595375 

18.3030 

6.9451 

291   84681 

24642171J  17.0587 

6.6267 

336 

112896 

37933056 

18.3303 

6.9521 

292   85264 

24897088  17.0880 

6.6343 

337 

113569 

38272753 

18.3576 

6.9589 

293:  85849 

25153757  17.1172 

6.6419 

338 

114244 

38614472 

18.3848 

6.9658 

294   86436 

25412184  17.1464 

6.6494 

339 

114921 

38958219 

18.4120 

6.9727 

295 

87025 

25672375 

17.1756 

6.6569 

340 

115600 

39304000 

18.4391 

6.9795 

296 

87616 

25934336 

17.2047 

6.6644 

341 

116281 

39651821 

18.4662 

6.9864 

297   88209 

26198073 

17.2337 

6.6719 

342 

116964 

40001688  18.4932 

6.9932 

298   88804 

26463592 

17.2627 

6.6794 

343 

117649 

40353607  18.5203 

7. 

299   89401 

26730899 

17.2916 

6.6869 

344 

118336 

40707584 

18.5472 

7.0068 

300   90000 

27000000 

17.3205 

6.6943 

345 

119025 

41063625 

18.5742 

7.0136 

301 

90601 

27270901 

17.3494 

6.7018 

346 

119716 

41421736 

18.6011 

7.0203 

302 

91204   27543608 

17.3781 

6.7092 

347 

120409 

41781923  18.6279 

7.0271 

303;  91809   27818127 

17.4069 

6.7166 

348 

121104 

42144192!  18.6548 

7.0338 

304   92416   28094464 

17.4356 

6.7240 

349 

121801 

42508549 

18.6815 

7.0406 

305  93025 

28372625 

17.4642 

6.7313 

350 

122500 

42875000 

18.7083 

7.0473 

306 

93636 

28652616 

17.4929 

6.7387 

351 

123201 

43243551 

18.7350 

7.0540 

307   94249   28934443 

17.5214 

6.7460 

352 

123904 

43614208 

18.7617 

7.0607 

308   94864  29218112 

17.5499 

6.7533 

353 

124609 

43986977 

18.7883 

7.0674 

309   95481 

29503629 

17.5784 

6.7606 

354 

125316 

44361864 

18.8149 

7.0740 

310 

96100 

29791000 

17.6068 

6.7679 

355 

126025 

44738875 

18.8414 

7.0807 

311   96721 

30080231 

17.6352 

6.7752 

356 

126736 

45118016 

18.8680 

7.0873 

312 

97344 

30371328 

17.6635 

6.7824 

357 

127449 

45499293 

18.8944 

7.0940 

313 

97969   30664297 

17.6918 

6.7897  358 

128164 

45882712;  18.9209 

7.1006 

314 

98596   30959144 

17.7200 

6.7969  3.59 

128881 

46268279  '  18.9473 

7.1072 

315 

99225 

31255875 

17.7482 

6.8041 

360 

129600 

46656000  18.9737   7.1138 

MATHEMATICAL  TABLES 


107 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
NUMBERS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

361 

130321 

47045881 

19. 

7.1204 

406 

164836 

66923416 

20.1494 

7.4047 

362 

131044 

47437928  19.0263 

7.1269; 

407  165649   67419143  20.1742 

7.4108 

363 

131769   47832147  19.0526   7.1335| 

408!  166464   67917312  20.1990 

7.4169 

364 

132196   48228544  19.0788   7.1400 

409 

167281]  68417929  20.2237 

7.4229 

365 

133225 

48627125 

19.1050 

7.1466 

410 

168100 

68921000  20.2485 

7.4290 

366 

133956 

49027896'  19.1311 

7.1531 

411 

168921 

69426531 

20.2731 

7.4350 

367 

134689 

49430863  19.1572 

7.1596 

412  169744   69934528  20.2978 

7.4410 

368 

135424 

49836032  19.1833 

7.1661 

413  170569 

70444997  20.3224 

7.4470 

369 

136161 

50243409  19.2094 

7.1726 

414 

171396 

70957944  '•  20.3470 

7.4530 

370 

136900 

50653000  19.2354 

7.1791 

415 

172225 

71473375  20.3715 

7.4590 

371 

137641 

51064811 

19.2614 

7.1855 

416 

173056 

71991296  20.3961 

7.4650 

372 

138384 

51478848  19.2873 

7.1920 

417 

173889 

72511713,  20.4206 

7.4710 

373 

139129 

51895117  19.3132 

7.1984  418 

174724 

73034632;  20.4450 

7.4770 

374 

139876 

52313624  19.3391 

7.2048 

419  175561 

73560059:  20.4695 

7.4829 

375 

140625 

52734375 

19.3649 

7.2112 

420 

176400 

74088000 

20.4939 

7.4889 

376 

141376 

53157376 

19.3907 

7.2177 

421 

177241 

74618461 

20.5183 

7.4948 

377 

142129 

53582633  19.4165   7.2240 

422  178084   75151448 

20.5426 

7.5007 

378 

142884 

51010152  19.4422   7.2304 

423  178929 

75686967 

20.5670 

7.5067 

379 

143641 

54439939  19.4679 

7.2368 

424 

179776 

76225024 

20.5913 

7.5126 

380 

144400 

54872000 

19.4936 

7.2432 

425 

180625 

76765625 

20.6155 

7.5185 

381 

145161 

55306341  19.5192 

7.2495 

426 

181476 

77308776 

20.6398 

7.5244 

382 

145924 

55742968  19.5448   7.2558 

427 

182329 

77854483  20.6640 

7.5302 

383 

146689 

56181887  19.5704 

7.2622 

428 

183184 

78402752 

20.6882 

7.5361 

384 

147456 

56623104  19.5959 

7.2685 

429 

184041 

78953589 

20.7123 

7.5420 

385 

148225 

57066625 

19.6214 

7.2748 

430 

184900 

79507000 

20.7364 

7.5478 

386 

148996 

57512456 

19.6469 

7.2811 

431 

185761 

80062991 

20.7605 

7.5537 

387 

149769 

57960603  19.6723 

7.2874  432 

186624 

80621568;  20.7846 

7.5595 

388 

150544 

58411072  19.6977 

7.2936  433 

187489 

81182737  20.8087 

7.5654 

389 

151321 

58863869  19.7231 

7.  2999  '  434 

188356   81746504  20.8327 

7.5712 

390 

152100 

59319000  19.7484 

7.3061 

435 

189225 

82312875 

20.8567 

7.5770 

391 

152881 

59776471 

19.7737 

7.3124 

436 

190096 

82881856 

20.8806 

7  .  5828 

392 

153664   60236288  19.7990 

7.3186 

437 

190969 

83453453  20.9045 

7.5886 

393 

154449   60698457  19.8242 

7.3248 

438 

191884 

84027672  20.9284 

7.5944 

391 

155236   61162984  19.8494 

7.3310 

439  192721 

84604519 

20.9523 

7.6001 

395 

156025 

61629875 

19.8746 

7.3372 

440 

193600 

85184000 

20.9762 

7.6059 

396 

156816   62099136  19.8997 

7.3434 

441 

194481 

85766121 

21. 

7.6117 

397 

157609   62570773  19.9249 

7.3496 

442 

195364 

86350888  21.0238 

7.6174 

398 

158404   63044792 

19.9499   7.3558 

443 

196249 

86938307  1  21.0476 

7.6232 

399 

159201   63521199  19.9750   7.3619 

444 

197136 

87528384  21.0713 

7.6289 

400 

160000   64000000 

20. 

7.3681 

445 

198025 

88121125 

21.0950 

7.6346 

401 

160801   64481201 

20.0250 

7.3742 

446 

198916 

88716536  21.1187 

7.6403 

402 

161604   64964808  20.0499   7.3803 

447 

199809   89314623  21.1424 

7.6460 

403 

162409 

65450827  20.0749   7.3864 

448 

200704 

89915392  21.1660 

7.6517 

404 

163216 

65939264!  20.0998 

7.3925 

449 

201601 

90518S49  21.1896 

7.6574 

405 

164025 

66430125 

20.1246 

7.3986 

450 

202500 

91125000 

21.2132 

7.6631 

108 


SHOP  ARITHMETIC 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
NUMBERS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

451 

203401 

91733851 

21.2368 

7.6688 

496 

246016 

122023936 

22.2711 

7.9158 

452 

204304 

92345408 

21.2603 

7.6744 

497 

247009 

122763473 

22.2935 

7.9211 

453 

205209 

92959677 

21.2838 

7.6801 

498 

248004 

123505992  22.3159 

7.9264 

454 

206116 

93576664 

21.3073 

7.6857 

499 

249001 

124251499 

2-2.3383 

7.9317 

455 

207025 

94196375 

21.3307 

7.6914 

500 

250000 

125000000 

22.3607 

7.9370 

456 

207936 

94818816 

21.3542 

7.6970 

501 

251001 

125751501 

22.3830 

7.9423 

457  208849 

95443993  21.3776 

7.7026 

502 

252004 

126506008  22.4054 

7.9476 

458  209764 

96071912  21.4009 

7.7082 

503 

253009 

127263527  22.4277 

7.9528 

459 

210681 

96702579  21.4243 

7.7138 

504 

254016 

128024064  22.4499 

7.9581 

460 

211600 

97336000 

21.4476 

7.7194 

505 

255025 

128787625 

22.4722 

7.9634 

461 

212521 

97972181 

21.4709 

7.72.50 

506 

256036 

129554216 

22.4944 

7.9686 

462 

213444 

98611128 

21.4942 

7.7306 

507  257049 

130323843  22.5167 

7.9739 

463 

214369 

99252847 

21.5174 

7.7362 

508 

258064 

131096512  22.5389 

7.9791 

464 

215296 

99897344 

21.5407 

7.7418 

509 

259081 

131872229 

22.5610   7.9843 

465 

216225 

100544625 

21.5639 

7.7473 

510 

260100 

132651000 

22.5832 

7.9896 

466 

217156 

101194696 

21.5870 

7.7529 

511 

261121 

133432831 

22.6053 

7.9948 

467 

218089 

101847563 

21.6102 

7.7584 

512 

262144 

134217728 

22.6274   8. 

468 

219024 

102503232 

21.6333 

7.7639 

513 

263169 

135005697 

22.6495  j  8.0052 

469 

219961 

103161709 

21.6564 

7.7695 

514 

264196 

135796744 

22.6716   8.0104 

470 

220900 

103823000 

21.6795 

7.7750 

515 

265225 

136590875 

22.6936   8.0156 

471 

221841 

104487111 

21.7025 

7.7805 

516 

266256 

137388096 

22.7156   8.0208 

472  222784 

105154048 

21.7256 

7.7860, 

517 

2672S9 

138188413  22.7376   8.0260 

473 

223729 

105823817 

21.7486 

7.7915 

518 

268324 

138991832  22.7596   8.0311 

474 

224676 

106496424  21.7715 

7.7970 

519 

269361 

139798359 

22.7816   8.0363 

475 

225625 

107171875 

21.7945 

7.8025 

520 

270400 

140608000 

22.8035 

8.0415 

476 

226576 

107850176 

21.8174 

7.8079 

521 

271441 

141420761 

22.8254 

8.0466 

477 

227529 

108531333  21.8403 

7.8134 

522 

272484 

142236648  22.8473   8.0517 

478 

228484 

109215352  21.8632 

7.8188 

523 

273529 

143055667  22.8692   8.0569 

479 

229441 

109902239  21.8861 

7.8243 

524 

274576 

143877824  22.8910   8.0620 

480 

230400 

110592000 

21.9089 

7.8297 

525 

275625 

144703125 

22.9129 

8.0671 

481 

231361 

111284641 

21.9317 

7.8352 

526 

276676 

145531576 

22.9347 

8.0723 

482 

232324 

111980168  21.9545   7.8406 

527 

277729 

146363183  22.9565   8.0774 

483 

233289 

112678587  21.9773 

7.8460 

528 

278784 

147197952  22.9783 

8.0825 

484 

234256 

113379904 

22. 

7.8514 

529 

279841 

148035889  23. 

8.0876 

485 

235225 

114084*25 

22.0227 

7.8568 

530 

280900 

148877000 

23.0217 

8.0927 

486 

236196 

114791256 

22.0454 

7.8622 

531 

281961 

149721291 

23.0434 

8.0978 

487 

237169 

115501303  22.0681 

7.8676 

532 

283024 

150568768  23.0651   8.1028 

488 

238144 

116214272  22.0970 

7.8730 

533 

284089 

151419437  23.0868   8.1079 

489 

239121 

116930169  22.1133 

7.8784 

534 

285156 

152273304  23.1084 

8.1130 

490 

240100 

117649000 

22.1359 

7.8837 

535 

286225 

153130375 

23.1301 

8.1180 

491 

241081 

118370771 

22.1585 

7.8891 

536 

287296 

153990656 

23.1517 

8.1231 

492 

242064 

119095488 

22.1811 

7.8944 

537 

288369 

154854153  23.1733 

8.1281 

493 

243049 

119823157 

22.2036 

7.8998 

538 

289444 

155720872  23.1948 

8.1332 

494 

244036 

120553784 

22.2261 

7.9051 

539 

290521 

156590819  23.2164 

8.1382 

495 

245025 

121287375 

22.2486 

7.9105 

540 

291600 

157464000  23.2379 

8.1433 

MATHEMATICAL  TABLES 


109 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
N  UM  BE  RS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

541 

292681 

158340421 

23.2594 

8.1483 

586 

343396 

201230056 

24.2074 

8.3682 

642 

293764 

159220088 

23.2809 

8.1533 

587 

344569  202262003 

24.2281 

8.3730 

543 

294849 

160103007  23.3024 

8.1583 

588 

345744  203297472 

24.2487 

8.3777 

544 

295936 

160989184  23.3238 

8.1633 

589 

346921  204336469 

24.2693 

8.3825 

545 

297025 

161878625 

23.3452 

8.1683 

590 

348100 

205379000 

24.2899 

8.3872 

r>  10 

298116 

162771336 

23.3666 

8.1733 

591 

349281 

206425071 

24.3105 

8.3919 

547  299209 

163667323  23.3880 

8.1783 

592 

350464 

207474688 

24.3311 

8.3967 

548  300304 

164566592  23.4094 

8.1833 

593 

351649  208527857 

24.3516 

8.4014 

549  301401 

165469149  23.4307 

8.1882 

594 

352836 

209584584 

24.3721 

8.4061 

550 

302500 

166375000 

23.4521 

8.1932 

595 

354025 

210644875 

24.3926 

8.4108 

551 

303601 

167284151 

23.4734 

8.1982 

596 

355216 

211708736 

24.4131 

8.4155 

552  304704  168196608 

23.4947 

8.2031 

597 

356409 

212776173 

24.4336 

8.4202 

553  305809  169112377 

23.5160 

8.2081 

598 

357604 

213847192 

24.4540 

8.4249 

554  306916  170031464 

23.5372 

8.2130 

599 

358801 

214921799 

24.4745 

8.4296 

555 

308025 

170953875 

23.5584 

8.2180 

600 

360000 

216000000 

24.4949 

8.4343 

556 

309136 

171879616 

23.5797 

8.2229 

601 

361201 

217081801 

24.5153 

8.4390 

557 

310249 

172808693  23.6008 

8.2278 

602 

362404 

218167208 

24.5357 

8.4437 

558  311364  173741112  23.6220 

8.2327 

603 

363609 

219256227 

24.5561 

8.4484 

559  312481  174676879 

23.6432 

8.2377 

604 

364816 

220348864 

24.5764 

8.4530 

560  313600 

175616000 

23.6643 

8.2426 

605 

366025 

221445125 

24.5967 

8.4577 

561  314721 

176558481 

23.6854 

8.2475 

606 

367236 

222545016 

24.6171 

8.4623 

562  315844 

177504328 

23.7065 

8.2524 

607 

368449 

223648543 

24.6374 

8.4670 

563  316969 

178453547 

23.7276 

8.2573 

608 

369664 

224755712 

24.6577 

8.4716 

564  318096 

179406144 

23.7487 

8.2621 

609 

370881 

225866529 

24.6779 

8.4763 

56.') 

319225 

180362125 

23.7697 

8.2670 

610 

372100 

226981000 

24.6982 

8.4809 

566 

320356 

181321496 

23.7908 

8.2719 

611 

373321 

228099131 

24.7184 

8.4856 

567 

321489  182284263 

23.8118 

8.2768 

612 

374544 

229220928 

24.7386 

8.4902 

568 

322624  183250432 

23.8328 

8.2816 

613 

375769 

230346397 

24.7588 

8.4948 

569 

323761  ]  184220009 

23.8537 

8.2865 

614 

376996 

231475544 

24.7790 

8.4994 

570 

324900 

185193000 

23.8747 

8.2913 

615 

378225 

232608375 

24.7992 

8.5040 

571 

326041 

186169411 

23.8956 

8.2962 

616 

379456 

233744896 

24.8193 

8.5086 

672 

327184 

187149248 

23.9165 

8.3010 

617 

380689 

234885113 

24.8395 

8.5132 

573 

328329  188132517 

23.9374 

8.3059 

618 

381924 

236029032 

24.8596 

8.5178 

571 

329476,  189119224 

23.9583 

8.3107 

619 

383161 

237176659 

24.8797 

8.5224 

575 

330625 

190109375 

23.9792 

8.3155 

620 

384400 

238328000 

24.8998 

8.5270 

576 

331776 

191102976 

24. 

8.3203 

621 

385641 

239483061 

24.9199 

8.5316 

577 

332929 

192100033 

24.0208 

8.3251 

622 

386884 

240641848 

24.9399 

8.5362 

578 

334084 

193100552 

24.0416 

8.3300 

623 

388129 

241804367 

24.9600 

8.5408 

579 

335241 

194104539 

24.0624 

8.3348 

624 

389376 

242970624 

24.9800 

8.5453 

580 

336400 

195112000 

24.0832 

8.3396 

625 

390625 

244140625 

25. 

8.5499 

581 

337561 

196122941 

24.1039 

8.3443 

626 

391876 

245314376 

25.0200 

8.5544 

582 

338724  197137368 

24.1247 

8.3491 

627 

393129 

246491883 

25.0400 

8.5590 

583 

339889;  198155287 

24.1454 

8.3539 

628 

394384 

247673152 

25.0599 

8.5635 

584 

341056 

199176704 

24.1661 

8.3587 

629 

395641 

248858189 

25.0799 

8.5681 

585 

342225 

200201625 

24.1868 

8.3634 

630 

396900 

250047000 

25.0998 

8.5726 

110 


SHOP  ARITHMETIC 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
NUMBERS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

631 

398161 

251239591 

25.1197 

8.5772 

676 

456976 

308915776 

26. 

8.7764 

632 

399424 

252435968 

25.1396 

8.5817 

677 

458329 

310288733 

26.0192 

8.7807 

633 

400689 

253636137 

25.1595 

8.5862 

678 

459684 

311665752 

26.0384 

8.7850 

634 

401956 

254840104 

25.1794 

8.5907 

679 

461041 

313046839 

26.0576 

8.7893 

635 

403225 

256047875 

25.1992 

8.5952 

680 

462400 

314432000 

26.0768 

8.7937 

636 

404496 

257259456 

25.2190 

8.5997 

681 

463761 

315821241 

26.0960 

8.7980 

637 

405769 

258474853 

25.2389 

8.6043 

682 

465124 

317214568 

26.1151 

8.8023 

638 

407044 

259694072 

25.2587 

8.6088 

683 

466489 

318611987 

26.1343 

8.8066 

639 

408321 

260917119 

25.2784 

8.6132 

684 

467856 

320013504 

26.1534 

8.8109 

640 

409600 

262144000 

25.2982 

8.6177 

685 

469225 

321419125 

26.1725 

8.8152 

641 

410881 

263374721 

25.3180 

8.6222 

686 

470596 

322828856 

26.1916 

8.8194 

642 

412164 

264609288 

25.3377 

8.6267 

687 

471969 

324242703 

26.2107 

8.8237 

643 

413449 

265847707 

25.3574 

8.6312 

688 

473344 

325660672 

26.2298 

8.8280 

644 

414736 

267089984 

25.3772 

8.6357 

689 

474721 

327082769 

26.2488 

8.8323 

645 

416025 

268336125 

25.3969 

8.6401 

690 

476100 

328509000 

26.2679 

8.8366 

646 

417316 

269586136 

25.4165 

8.6446 

691 

477481 

329929371 

26.2869 

8.8408 

647 

418609 

270840023 

25.4362 

8.6490 

692 

478864 

331373888 

26.3059 

8.8451 

648 

419904 

272097792 

25.4558 

8.6535 

693 

480249 

332812557 

26.3249 

8.8493 

649 

421201 

273359449 

25.4755 

8.6579 

694 

481636 

334255384 

26.3439 

8.8536 

650 

422500 

274625000 

25.4951 

8.6624 

695 

483025 

335702375 

26.3629 

8.8578 

651 

423801 

275894451 

25.5147 

8.6668 

696 

484416 

337153536 

26.3818 

8.8621 

652 

425104 

277167808 

25.5343 

8.6713 

697 

485809 

338608873 

26.4008 

8.8663 

653 

426409 

278445077 

25.5539 

8.6757 

698 

487204 

340068392 

26.4197 

8.8706 

654 

427716 

279726264 

25.5734 

8.6801 

699 

488601 

341532099 

26.4386 

8.8748 

655 

429025 

2S1011375 

25.5930 

8.6845 

700 

490000 

343000000 

26.4575 

8.8790 

656 

430336 

282300416 

25.6125 

8.6890 

701 

491401 

344472101 

26.4764 

8.8833 

657 

431649 

283593393 

25.6320 

8.6934 

702 

492804 

345948408 

'26.4953 

8.8875 

658 

432964 

284890312 

25.6515 

8.6978 

703 

494209 

347428927 

26.5141 

8.8917 

659 

434281 

286191179 

25.6710 

8.7022 

704 

495616 

348913664 

26.5330 

8.8959 

660 

435600 

287496000 

25.6905 

8.7066 

705 

497025 

350402625 

26.5518 

8.9001 

661 

436921 

288804781 

25.7099 

8.7110 

706 

498436 

351895816 

26.5707 

8.9043 

662 

438244 

290117528 

25.7294 

8.7154 

707 

499849 

353393243 

26.5895 

8.9085 

663 

439569 

291434247 

25.7488 

8.7198 

708 

501264 

354894912 

26.6083 

8.9127 

664 

440896 

292754944 

25.7682 

8.7241 

709 

502681 

356400829 

26.6271 

8.9169 

665 

442225 

294079625 

25.7876 

8.7285 

710 

504100 

357911000 

26.6458 

8.9211 

666 

443556 

295408296 

25.8070 

8.7329 

711 

505521 

359425431 

26.6646 

8.9253 

667 

444889 

296740963 

25.8263 

8.7373 

712 

506944 

360944128 

26.6833 

8.9295 

668 

446224 

298077632 

25.8457 

8.7416 

713 

508369 

362467097 

26.7021 

8.9337 

669 

447561 

299418309 

25.8650 

8.7460 

714 

509796 

363994344 

26.7208 

8.9378 

670 

448900 

300763000 

25.8844 

8.7503 

715 

511225 

365525875 

26.7395 

8.9420 

671 

450241 

302111711 

25.9037 

8.7547 

716 

512656 

367061696 

26.7582 

8.9462 

672 

451584 

303464448 

25.9230 

8.7590 

717 

514089 

368601813 

26  .  7769 

8.9503 

673 

452929 

304821217 

25.9422 

8.7634 

718 

515524 

370146232 

26.7955 

8.9545 

674 

454276 

306182024 

25.9615 

8.7677 

719 

516961 

371694959 

26.8142 

8.9587 

675 

455625 

307546875 

25.9808 

8.7721 

720 

518400 

373248000 

26.8328 

8.9628 

MATHEMATICAL  TABLES 


111 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
NUMBERS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

721 

519841 

374805361  26.8514 

8.9670 

766 

586756 

449455096 

27.6767 

9.1498 

722 

521284  376367048  26.8701 

8.9711 

767 

588289  451217663  27.6948 

9.1537 

723 

522729  377933067  26.8887 

8.9752 

768 

589824  452984832  27.7128 

9.1577 

724 

524176 

379503424  26.9072 

8.9794 

769 

591361 

454756609  27.7308 

9.1617 

725 

525625 

381078125 

26.9258 

8.9835 

770 

592900 

456533000 

27.7489 

9.1657 

726 

527076 

382657176 

26.9444 

8.9876 

771 

594441 

458314011 

27.7669 

9.1696 

727 

528529  384240583  26.9629 

8.9918 

772 

595984 

460099648  27.7849 

9.1736 

728 

529984  385828352 

26.9815 

8.9959 

773 

597529 

461889917  27.8029 

9.1775 

729 

531441  387420489  27. 

9. 

774 

599076 

463684824 

27.8209 

9.1815 

730 

532900 

389017000 

27.0185 

9.0041 

775 

600625 

465484375 

27.8388 

9.1855 

731 

534361 

390617891 

27.0370 

9.0082 

776 

602176 

467288576 

27.8568 

9.1894 

732 

535824 

392223168J  27.0555 

9.0123 

777 

603729 

469097433 

27.8747 

9.1933 

733 

537289 

393832837  27.0740 

9.0164 

778 

605284 

470910952 

27.8927 

9.1973 

734 

538756 

395446904 

27.0924 

9.0205 

779 

606841 

472729139 

27.9106 

9.2012 

735 

540225 

397065375 

27.1109 

9.0246 

780 

608400 

474552000 

27.9285 

9.2052 

736 

541696 

398688256 

27.1293 

9.0287 

781 

609961 

476379541 

27.9464 

9.2091 

737 

543169 

400315553 

27.1477 

9.0328 

782 

611524 

478211768  27.9643 

9.2130 

738 

544644 

401947272  27.1662 

9.0369 

783 

613089 

480048687;  27.9821 

9.2170 

739 

546121 

403583419  27.1846 

9.0410 

784 

614656 

481890304 

28. 

9.2209 

740 

547600 

405224000 

27.2029 

9.0450 

785 

616225 

483736625 

28.0179 

9.2248 

741 

549081 

406869021  27.2213 

9.0491 

786 

617796 

485587656 

28.0357 

9.2287 

742 

5505641  408518488 

27.2397 

9.0532 

787  619369  487443403 

28.0535 

9.2326 

743 

552049 

410172407 

27.2580 

9.0572 

788  620944  489303872 

28.0713 

9.2365 

744 

553536 

411830784 

27.2764 

9.0613 

789 

622521;  491169069 

28.0891 

9.2404 

745 

555025 

413493625 

27.2.947 

9.0654 

790 

624100 

493039000 

28.1069 

9.2443 

746 

556516 

415160936 

27.3130 

9.0694 

791 

625681 

494913671 

28.1247 

9.2482 

747 

558009 

416832723 

27.3313 

9.0735 

792 

627264 

496793088  28.1425 

9.2521 

748 

559504 

418508992 

27.3496 

9.0775 

793 

628849 

498677257  28.1603 

9.2560 

749 

561001 

420189749 

27.3679 

9.0816 

794 

630436 

500566184  28.1780 

9.2599 

750 

562500 

421875000 

27.3861 

9.0856 

795 

632025 

502459875 

28.1957 

9.2638 

751 

564001 

423564751 

27.4044 

9.0896 

796 

633616 

504358336 

28.2135 

9.2677 

752 

565504  425259008 

27.4226 

9.0937 

797  635209  506261573  28.2312 

9.2716 

753 

567009 

426957777 

27.4408 

9.0977 

798  636804  508169592  28.2489 

9.2754 

754 

568516 

428661064 

27.4591 

9.1017 

799  638401  510082399  28.2666 

9.2793 

755 

570025 

430368875 

27.4773 

9.1057 

800 

640000 

512000000  28.2843 

9.2832 

756 

571536 

432081216 

27.4955 

9.1098 

801 

641601 

513922401 

28.3019 

9.2870 

757 

573049  433798093 

27.5136 

9.1138 

802  643204 

5158496081  28.3196 

9.2909 

758 

574564  435519512 

27.5318   9.1178 

803:  644809 

517781627  28.3373 

9.2948 

759 

576081  437245479 

27.5500  9.1218 

804  646416 

519718464  28.3549 

9.2986 

760 

577600 

438976000 

27.5681 

9.1258 

805 

648025 

521660125 

28.3725 

9.3025 

761 

579121 

440711081 

27.5862 

9.1298 

806  649636 

523606616  28.3901 

9.3063 

762 

580644  442450728  27.6043 

9.1338 

807  651249;  525557943  28.4077 

9.3102 

763 

-582169  444194947  27.6225 

9.1378 

808  652864  527514112  28.4253   9.3140 

764 

583696;  445943744  27.6405 

9.1418 

809  654481 

529475129  28.4429   9.3179 

765 

585225  447697125  27.6586   9.1458 

810  656100 

531441000  28.4605   9.3217 

112 


SHOP  ARITHMETIC 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
N  UM  BERS .— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

811 

657721 

533411731 

28.4781 

9.3255 

856 

732736 

627222016 

29.2575 

9.4949 

812 

659344 

535387328 

28.4956 

9.3294 

857 

734449 

629422793  29.2746 

9.4986 

813 

660969 

537367797 

28.5132 

9.3332 

858  736164  631628712  29.2916 

9.5023 

814 

662596 

539353144 

28.5307 

9.3370  859  737881'  633839779  29.3087 

9.5060 

815 

664225 

541343375 

28.5482 

9.3408 

860 

739600 

636056000 

29.3258 

9.5097 

816 

665856 

543338496 

28.5657 

9.3447 

861 

741321 

638277381 

29.3428 

9.5134 

817 

667489 

545338513 

28.5832 

9.3485 

862  743044 

640503928 

29.3598 

9.5171 

818 

669124 

547343432 

28.6007 

9.3523 

863  744769 

642735647 

29.3769 

9.5207 

819 

670761 

549353259 

28.6182 

9.3561 

864  746496  644972544 

29.3939 

9.5244 

820 

672400 

551368000 

28.6356 

9.3599 

865 

748225 

647214625 

29.4109 

9.5281 

821 

674041 

553387661 

28.6531 

9.3637 

866 

749956 

649461896 

29.4279 

9.5317 

822!  675684 

555412248 

28.6705 

9.3675 

867  751689T  651714363 

29.4449 

9.5354 

823 

677329 

557441767 

28.6880 

9.3713 

868  753424 

653972032 

29.4618 

9.5391 

824 

678976 

559476224 

28.7054 

9.3751 

869  755161 

656234909 

29.4788 

9.5427 

825 

680625 

561515625 

28.7228 

9.3789 

870 

756900 

658503000 

29.4958 

9.5464 

826 

682276 

563559976 

28.7402 

9.3827 

871 

758641 

660776311 

29.5127 

9.5501 

827 

683929 

565609283 

28.7576 

9.3865 

872  760384 

663054848 

29.5296 

9.5537 

828 

685584 

567663552  28.7750 

9.3902  873 

762129 

665338617 

29.5466 

9.5574 

829 

687241 

569722789 

28.7924 

9.3940 

874 

763876 

667627624 

29.5635 

9.5610 

830 

688900 

571787000 

28.8097 

9.3978 

875 

765625 

669921875 

29.5804 

9.5647 

831 

690561 

573856191 

28.8271 

9.4016 

876 

767376 

672221376 

29.5973 

9.5683 

832 

692224 

575930368 

28.8444 

9.4053 

877 

769129 

674526133 

29.6142 

9.5719 

833 

693889  578009537 

28.8617 

9.4091 

878 

770884 

676836152 

29.6311 

9.5756 

834 

695556 

580093704 

28.8791 

9.4129 

879 

772641 

679151439 

29.6479 

9.5792 

835 

697225 

582182875 

28.8964 

9.4166 

880 

774400 

681472000 

29.6648 

9.5828 

836 

698896 

584277056 

28.9137 

9.4204 

881 

776161 

683797841 

29.6816 

9.5865 

837 

700569 

586376253 

28.9310 

9.4241 

882 

777924 

686128968 

29.6985 

9.5901 

838 

702244 

588480472 

28.9482 

9.4279 

883 

779689 

688465387 

29.7153 

9.5937 

839 

703921 

590589719 

28.9655 

9.4316 

884 

781456 

690807104 

29.7321 

9.5973 

840 

705600 

592704000 

28.9828 

9.4354 

885 

783225 

693154125 

29.7489 

9.6010 

841 

707281 

594823321 

29. 

9.4391 

886 

784996 

695506456 

29.7658 

9.6046 

842  708964 

596947688 

29.0172 

9.4429 

887 

786769 

697864103 

29  .  7825 

9.6082 

843  710649 

599077107 

29.0345 

9.4466 

888 

788544 

700227072 

29.7993 

9.6118 

844 

712336 

601211584 

29.0517 

9.4503 

889 

790321 

702595369 

29.8161 

9.6154 

845 

714025 

603351125 

29.0689 

9.4541 

890 

792100 

704969000 

29.8329 

9.6190 

846 

715716 

605495736 

29.0861 

9.4578 

891 

793881 

707347971 

29.8496 

9.6226 

847 

717409 

607645423 

29.1033 

9.4615 

892 

795664 

709732288 

29.8664 

9.6262 

848 

719104 

609800192 

29.1204 

9.4652 

893 

797449 

712121957  29.8831 

9.6298 

849 

720801 

611960049 

29  .  1376 

9.4690 

894 

799236 

714516984  29.8998 

9.6334 

850 

722500 

614125000 

29.1548 

9.4727 

895 

801025 

716917375  29.9166 

9.6370 

851 

724201 

616295051 

29.1719 

9.4764 

896 

802816 

719323136  29.9333 

9.6406 

852 

725904 

618470208  29.1890 

9.4801 

897  804609  721734273  29.9500   9.6142 

853 

727609 

6206.50477  29.2062 

9.4838 

898 

806  10  i  724150792  29.9666   9.6477 

854 

729316 

622835864 

29.2233 

9.4875 

899 

808201  726572699  29.9833 

9.6513 

855 

731025 

625026375 

29.2404 

9.4912 

900 

810000-  729000000  30. 

9.6549 

MATHEMATICAL  TABLES 


113 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
NUMBERS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

901 

811801 

731432701 

30.0167 

9.6585 

946 

894916 

846590536 

30.7571 

9.8167 

902  813604 

733870808 

30.0333   9.6620  947 

896809 

849278123  30.7734 

9.8201 

903  815409 

736314327 

30.0500   9.6656  948  898704 

851971392  30.7896 

9.8236 

901  817216 

738763264 

30.0666   9.6692 

949  900601 

854670349  30.8058 

9.8270 

903  819025 

741217625 

30.0832 

9.6727 

950  902500 

857375000  30.8221 

9.8305 

906  820836 

743677416 

30.0998 

9.6763 

951 

904401 

860085351 

30.8383 

9.8339 

907  822619,  746142643 

30.1164   9.6799  952  906304 

862801408  30.8545 

9.8374 

908  824461 

748613312 

30.1330   9.6834  953  908209  865523177  30.8707 

9.8408 

909  826281 

751089429 

30.1496   9.6870  954  910116  86S250664  30.8869 

9.8443 

910  828100 

753571000 

30.1662 

9.6905 

955 

912025 

870983875  30.9031 

9.8477 

911  829921 

756058031 

30.1828 

9.6941 

956 

913936 

873722816  30.9192 

9.8511 

912  831744 

758550528 

30.1993 

9.6976  957 

915849 

876467493  30.9354 

9.8546 

913  833569 

761048497 

30.2159 

9.7012  958 

917764|  879217912  30.9516 

9.8580 

914  835396 

763551944 

30.2324 

9.7047 

959 

919681  881974079  30.9677 

9.8614 

<Jlo 

837225 

766060875 

30.2490 

9.7082 

960 

921600 

884736000  30.9839 

9.8648 

916 

839056 

768575296 

30.2655 

9.7118 

961 

923521 

887503681 

31. 

9.8683 

917  840889 

771095213 

30.2820 

9.7153 

962 

925444 

890277128  31.0161 

9.8717 

918  842724 

773620632 

30.2985 

9.7188  903 

927369  893056347  31.0322 

9.8751 

919  844561 

776151559 

30.3150 

9.7224  964  929296  895841344  31.0483 

9.8785 

920  846400 

778688000 

30.3315 

9.7259 

965 

931225 

898632125 

31.0644 

9.8819 

921  848241 

781229961 

30.3480 

9.7294 

966 

933156  901428696  31.0805 

9.8854 

922  850084 

783777448 

30.3645 

9.7329 

967  935089  904231063  31.0966 

9.8888 

923  851929 

786330467  30.3809 

9.7364 

968  937024  907039232  31.1127 

9.8922 

921  853776 

788889024 

30.3974 

9.7400 

969  938961  909853209 

31.1288 

9.8956 

925 

855625 

791453125 

30.4138 

9.7435 

970  940900  912673000 

31.1448 

9.8990 

926 

857476 

794022776 

30.4302 

9.7470 

971  942841 

915498611 

31.1609 

9.9024 

927  859329 

796597983 

30.4467 

9.7505 

972  944784  918330048  31.1769 

9.9058 

928  861184 

799178752 

30.4631 

9.7540 

973  946729!  921167317  31.1929 

9.9092 

929  863041 

801765089 

30.4795 

9.7575 

974'  948676  924010424  31.2090 

9.9126 

930 

864900 

804357000 

30.4959 

9.7610 

975  950625.  926859375 

31.2250 

9.9160 

931 

866761 

806954491 

30.5123 

9.7645 

976 

952576 

929714176 

31.2410 

9.9194 

932 

868624 

809557568 

30.5287   9.7680 

977  954529 

932574833  31.2570 

9.9227 

933  870489 

812166237 

30.5450 

9.7715 

978  956484 

935441352  31.2730 

9.9261 

0:u  872356 

814780504 

30.5614 

9.7750 

979  958441  938313739 

31.2890 

9.9295 

935 

874225 

817400375 

30.5778 

9.7785 

980 

960400 

941192000 

31.3050 

9.9329 

936  876096 

820025856 

30.5941 

9.7819 

981 

962361  944076141 

31.3209 

9.9363 

937  877969 

822656953 

30.6105 

9.7854 

982 

964324 

946966168 

31.3369   9.9396 

938  879844 

825293672 

30.6268 

9.7889 

983 

966289 

949862087  31.3528   9.9430 

939  881721 

827936019 

30.6431 

9.7924 

984 

968256 

952763904  31.3688   9.9464 

940 

883600 

830584000 

30.6594 

9.7959 

085 

970225 

955671625 

31.3847 

9.9497 

911 

885481 

833237621 

30.6757 

9.7993 

986 

972196 

958585256  31.4006 

9.9531 

942  887364 

835896888 

30.6920 

9.8028  987  974169  961504803  31.4166   9.9565 

943 

889249 

838561807 

30.7083 

9.8063  988  976111  96-4430272  31.4325   9.9598 

944 

891138 

841232384 

30  .  7246 

9.8097  989  978121  967361669  31.4484   9.9632 

945 

893025 

843908625 

30.7409 

9.8132.  990 

980100  970299000  31.4643   9.9666 

114 


SHOP  ARITHMETIC 


SQUARES,  CUBES,  SQUARE  ROOTS,  AND  CUBE  ROOTS  OF 
N  U  M  BE  RS.— Continued 


No. 

Square 

Cube 

Square 
root 

Cube 
root 

No. 

Square 

Cube 

Square 
root 

Cube 
root 

991 

9S20S1 

973242271 

31.4802 

9.9699 

996 

992016 

988047936 

31.5595 

9.9866 

992    984064 

976191488 

31.4960 

9.9733 

997 

994009 

991026973 

31.5753 

9.9900 

993    986049    979146657    31.5119 

9.9766 

998 

996004 

994011992 

31.5911 

9.9933 

994 

988036    982107784    31.5278 

9.9800 

999 

998001 

997002999 

31.6070 

9.9967 

995 

990025 

985074875 

31.5436 

9.9833 

1000 

1000000 

1000000000 

31.6228 

• 

10. 

CHAPTER  XIII 
LEVERS 

90.  Types  of  Machines. — All  machines  consist  of  one  or  more 
of  the  three  fundamental  types  of  machines — the  Lever,  the 
Cord,  and  the  Inclined  Plane  or  Wedge.     Any  piece  of  mechanism 
can  be  proved  to  be  of  one  or  more  of  these  types.     Pulleys, 
gears,  and  cranks  will  be  shown  to  be  forms'  of  Levers;  belts  and 
chains  come  under  the  type  called  the  Cord;  while  screws,  worms, 
and  cams  are  forms  of  Inclined  Planes.     They  are  all  used  to 
transmit  power  from  one  place  to  another  and  to  modify  it,  as 
desired. 

91.  The  Lever. — The  lever  is  probably  the  most  used  and  the 
simplest  type  of  machine.     We  are  all  familiar  with  it  in  its 
simplest  forms,  such  as  crow  bars,  shears,  pliers,  tongs,  and  the 
numerous  simple  levers  found  on  machine  tools. 

A  lever  is  a  rigid  rod  or  bar  so  arranged  as  to  be  capable  of 
turning  about  a  fixed  point.  This  fixed  point  about  which  the 
lever  turns  is  called  the  Fulcrum.  In  Fig.  37  the  fulcrum  is 


Fia.  37. 

represented  by  the  small  triangular  block  F.  The  position  of 
this  fulcrum  determines  the  effect  which  the  force  P  applied  at 
one  end  has  toward  lifting  the  weight  W  at  the  other  end.  If 
F  is  close  to  W,  a  comparatively  small  force  P  may  be  able  to 
raise  the  weight  W,  but  if  F  is  moved  away  from  W  and  placed 
close  to  P,  then  a  greater  force  will  be  required  at  P.  If  F  is  in 
the  middle,  P  and  W  will  be  just  equal. 

In  every  lever  there  are  two  opposing  tendencies:  first,  that 
of  the  load  or  weight  W  tending  to  descend;  and  second,  that  of 
the  force  P  tending  to  raise  W.  The  ability  of  W  to  descend  or 
to  resist  being  lifted  depends  on  two  things — its  weight  and  its 
distance  from  the  fulcrum  F.  The  product  of  these  two  is  the 
measure  of  the  tendency  of  W  to  descend.  This  product  is 
10  115 


116  SHOP  ARITHMETIC 

called,  in  books  on  mechanics,  the  Moment.  Likewise,  the  force 
P  has  a  moment,  which  is  the  product  of  the  force  P  and  the 
distance  from  P  to  the  fulcrum  F.  If  the  force  and  the  weight 
just  balance  each  other,  their  moments  are  equal. 

The  length  from  P  to  F  is  called  the  force  arm  and  the  length 
from  W  to  F,  the  weight  arm.  Then,  for  balance,  we  have  the 
equation: 

Force  X  force  arm  =  Weight  X  weight  arm 
If  we  let  P  stand  for  the  force 

a    stand  for  the  force  arm 
W  stand  for  the  weight 
and      b    stand  for  the  weight  arm 

as  shown  in  Figs.  39,  40,  and  41,  we  will  have  the  formula 


Although  the  force  and  weight  are  really  balanced  when  this 
formula  is  fulfilled,  still  we  use  the  formula  for  calculating  the 
forces  necessary  to  lift  weights.  The  very  slightest  increase  in 
the  force  above  that  necessary  for  balance  will  cause  W  to  rise 
and,  therefore,  we  can  say  practically  that  P  will  lift  W  if 

p=WXb 
a 

If  it  is  the  length  a  that  is  wanted,  we  can  see  that  P  would 
lift  W  when 

Wxb 

a  =  —  ...    • 


14.'- 


FIG.  38. 

Example  : 

We  have  a  lever  14  ft.  long,  with  the  fulcrum  placed  2  ft.  from  the 
end,  as  shown  in  Fig.  38;  how  much  force  must  we  exert  to  lift  1800  lb.?  In 
this  problem  a  is  12  ft.,  b  is  2  ft.,  and  W  is  1800  lb. 


=  1800X2  = 
Then,  if  Pxl2(or  12  XP)  is  3600 
P  will  be  3600  -H!  2 
P  =  3600  -nl2  =  300  lb.,  Answer. 


LEVERS 


117 


It  will  be  seen  that  the  relation  between  force,  weight,  force 
arm,  and  weight  arm,  can  be  written  as  an  inverse  proportion. 

Force  :  weight    =  weight  arm  :  force  arm 

or 

P       :       W       =  6          :          a 

This  form  of  expressing  the  relation  is  not  generally  as  useful  as 
the  other  form,  PXa  =  WXb.  It  is  very  useful,  however,  in 
cases  where  neither  the  force  arm  nor  weight  arm  are  known. 

Example : 

If  a  man  wanted  to  lift  a  750  Ib.  weight  by  means  of  a  12  ft. 

timber  used  as  a  lever,  where  would  he  place  the  fulcrum  so  that  his  whole 

weight  of  150  Ib.  would  just  raise  it? 

Explanation:  The  total  length  of  the  timber 
(12  ft.)  is  the  sum  of  a  and  6  (see  Fig.  39).  We 
can  find  the  ratio  of  b  to  a  which  is  the  same 
as  P:W  and  reduces  to  1:5.  If  the  ratio  is 
1 : 5,  then  the  whole  length  is  6  parts  of  which 
o  is  5  parts  and  6,  1  part.  Hence  a  =  10  ft.  and 
6'=  2  ft.,  and  the  fulcrum  must  be  placed  2  ft. 
from  the  weight. 


P 

150 

150 

b 


W  =b 

750  =  6 
750  =  1 
a  =1 


=  £  of  12  =  10  ft. 
=iof  12  =  2  ft. 


92.  Three  Classes  of  Levers. — Levers  are  divided  into  there 
kinds  or  classes  according  to  the  relative  positions  of  the  force, 
fulcrum,  and  weight. 

Those  shown  so  far  are  of  the  first  class,  Fig.  39,  in  which  the 
fulcrum  is  between  the  force  and  the  weight.  The  weight  is 
lifted  by  pushing  down  at  P. 


FIQ.  39. 


W 


Fia.  40. 


In  the  second  class,  Fig.  40,  the  weight  is  between  the  fulcrum 
and  the  force,  and  the  weight  is  lifted  by  pulling  up  at  P. 

In  the  third  class,  Fig.  41,  the  force  P  is  between  the  weight 
and  the  fulcrum  and,  therefore,  P  must  be  greater  than  the  load 
that  it  lifts.  The  weight  is  lifted  by  an  upward  force  at  P. 


118 


SHOP  ARITHMETIC 


In  all  these  types  the  same  rule  holds  that : 

Force  X  force  arm  =  weight  X  weight  arm 

or 
p      x         a          =      W      X  b 


V 

/ 

F 

57 

h 

1  J 

FIQ.  41. 

Particular  attention  should  be  given  to  the  fact  that  the  force 
arm  and  weight  arm  are  always  measured  from  the  fulcruhi. 
In  levers  of  class  2,  the  force  arm  is  the  entire  length  of  the  lever. 
In  class  3,  the  force  arm  is  shorter  than  the  weight  arm.  This 
type  may  be  seen  on  the  safety  valves  of  many  boilers  and  is 
used  so  that  a  small  weight  can  balance  a  considerable  pressure 
at  P. 

Quite  often  there  appear  to  be  two  weights,  or  two  forces,  on  a 
lever,  and  it  is  difficult  to  decide  which  to  designate  as  the  force 
and  which  as  the  weight.  It  really  makes  no  difference  which 
we  call  the  force  and  which  the  weight;  the  relations  between 
them  would  be  the  same  in  any  case. 


\z- 


f 


n 

1  W  1 

—  L_+_J  

M 

a- 

u 

--                  in1  *. 

Fia.  42. 


93.  Compound  Levers. — We  frequently  meet  with  compound 
levers;  but  problems  concerning  them  are  easily  reduced  to 
repeated  cases  of  single  levers,  the  force  of  one  lever  correspond- 
ing to  the  weight  of  the  next,  etc.  To  illustrate  this  we  will  solve 
the  following  example: 


LEVERS  119 

Example  : 

We  wish  to  lift  8000  Ib.  with  a  compound  lever  as  shown  in  Fig.  42, 
the  first  one  being  10  ft.  long  with  the  weight  2  ft.  from  the  end;  the  second 
16  ft.  long  with  the  fulcrum  4  ft.  from  the  end;  what  will  be  the  necessary 
force,  P2? 

PXa=Wxb  Explanation:    Taking  the  first,   or  lower 

p  X10  =  8000X2  lever,  we  find  it  to  be  an  example  of  the 

16000  second  class.     W  has  a  weight  arm  6  of  2  ft. 

PI  =  —  TQ—  =  1600  Ib.  The  force  has  an  arm  equal  to  the  whole 

w      p  _  iron  lh  length  of  the  lever,  or  10  ft.     The  necessary 

P  v-     w  vfc  force  on  the  end  of  this  lever  we  find  to  be 

^ 


IK 

v19  —  ID. 


—  . 

6400  1  The  second  lever  must  pull  upward  through 

P2  =  -  =  533=  Ib.,  Answer,     the  connection  with  a  force  of  1600  Ib.     In 

other  words,  the  8000  Ib.  weight  on  the  first 

lever  is  equivalent  to  a  1600  Ib.  weight  on  the  short  end  of  the  second 
lever.  The  first  lever  pulls  downward  the  same  amount  that  the  second 
pulls  up,  or  Pl  =  W2.  Having  this  1600  as  the  weight,  we  find  that  a  force 
of  533J  Ib.  is  needed  on  the  end  of  the  second  lever. 


94.  Mechanical  Advantage. — The  ratio  of  the  weight  to  the 
force  is  often  called  the  Mechanical  Advantage  of  the  lever;  this 
ratio  is  equal  to  the  force  arm  -r-  the  weight  arm.  In  the  com- 
pound lever  of  Fig.  42  the  M.  A.  (mechanical  advantage)  of  the 
first  lever  equals  10  -f- 2  or  5;  of  the  second,  12-^4  or  3;  the  M.  A. 
of  a  compound  lever  is  equal  to  the  product  of  the  M.  A.  of  the 
separate  single  levers;  hence,  of  the  given  compound  lever  the 
M.  A.  is  5  X3  or  15.  This  means  that  a  1  Ib.  force  will  lift  15  Ib.; 
10  Ib.  will  lift  150;  or  100  Ib.  will  lift  1500.  The  force  multiplied 
by  the  M.  A.  gives  the  weight  that  can  be  lifted,  or  the  weight 
divided  by  the  M.  A.  gives  the  necessary  force.  In  the  case 
shown  in  Fig.  42,  the  mechanical  advantage  is  15  and  conse- 
quently the  necessary  force  is  8000 -7-15  =  533  £  Ib. 

If  the  mechanical  advantage  of  a  lever  is  10,  then  1  Ib.  will 
lift  10  Ib.,  or  800  Ib.  will  lift  8000  Ib.,  etc.;  but  it  must  be  remem- 
bered that  the  1  Ib.  or  the  800  Ib.  must  travel  10  times  as  far  as 
the  10  Ib.  or  the  8000  Ib. 

If  a  lever  has  a  mechanical  advantage  of  10,  the  force  must 
travel  10  times  as  far  as  it  lifts  the  weight,  and  consequently  a 
lever  effects  no  saving  in  work.  Work  is  the  product  of  force, 
or  weight,  times  the  distance  moved,  and  is  the  same  for  either 
end  of  the  lever.  It  is  similar  to  carrying  a  lot  of  castings  to  the 
top  floor  of  a  building.  If  I  carry  half  of  them  at  a  time,  I  must 
make  two  trips;  if  I  carry  one-fourth  of  them  at  a  time,  I  must 
make  four  trips.  The  lighter  the  load,  the  more  trips  I  must 
make.  The  work  done  is  the  same  whatever  way  I  carry  them 


120 


SHOP  ARITHMETIC 


and  is  equal  to  the  product  of  the  total  weight  times  the  height 
to  which  the  load  must  be  carried. 

95.  The  Wheel  and  Axle. — This  is  a  name  given  in  mechanics 
to  the  modification  of  levers  that  enables  them  to  be  rotated 
continuously.  Fig.  43  shows  the  principle  of  this:  By  wrapping 
a  belt  or  rope  around  each  of  the  two  circular  bodies,  we  find 
that  the  pulls  in  the  cords  are  in  inverse  proportion  to  the  radii 
of  the  circles.  A  little  consideration  shows  that  the  wheel  and 
axle  may  easily  be  studied  as  a  force,  P,  with  lever  arm  Rt  equal 
to  the  radius  of  the  wheel;  and  a  weight,  W,  with  a  lever  arm  r, 
equal  to  the  radius  of  the  axle.  Two  pulleys  on  a  countershaft 


FIG.  43. 


might  be  likened  to  a  lever  in  the  same  way.  The  belt  which 
drives  the  countershaft  furnishes  the  force  P.  The  radius  of  this 
pulley  is  the  force  arm.  The  radius  of  the  other  countershaft 
pulley,  which  transmits  the  power  to  the  machine,  is  the  weight 
arm  and  the  pull  in  this  belt  is  the  weight.  Gears  also  are  levers 
that  can  be  rotated  continuously.  The  simplest  example  of 
the  use  of  the  axle  is  probably  the  windlass,  which  we  see  used  for 
hoisting,  house-moving,  etc.  See  Figs.  48  and  50. 

A  geared  windlass,  such  as  shown  in  Fig.  50,  is  a  case  of  com- 
pound levers.  The  crank  and  pinion  form  the  first  lever  and  the 
load  on  the  gear  teeth  is  transmitted  to  the  teeth  of  the  larger 
gear  and  becomes  the  force  of  the  other  lever,  which  consists  of 
the  large  gear  and  the  drum.v 


LEVERS  121 

Example : 

A  geared  windlass,  such  as  shown  in  Fig.  50,  has  a  crank  20  in. 
long;  the  small  gear  is  6  in.  in  diameter,  the  large  gear  is  30  in.  in  diameter, 
and  the  diameter  of  the  drum  is  6  in. 

What  load  could  be  raised  by  a  man  exerting  a  force  of  25  Ib.  on  the  crank? 

6-r-2  =  3  in.,  radius  of  pinion 
20n-3  =  6§,  M.  A.  of  crank  and  pinion 
30-5-2  =  15  in.,  radius  of  gear 

6n-2  =  3  in.,  radius  of  drum 
15-:- 3  =  5,  M.  A.  of  gear  and  drum 
6jf  X5  =  33J  in.,  total  mechanical  advantage 
25X33^=833  +  Ib.,  Answer. 

The  solution  of  this  problem  might  be  shortened  by  writing  all  the  work  in  a 
single  equation: 

30 
25X20X-2"     25X20X15 

-e-y- 3^3"=833+  lb"  Answer- 

2X2 

If  we  do  not  know  the  sizes  of  the  gears  in  inches,  but  know  the 
numbers  of  teeth,  we  can  figure  that  the  mechanical  advantage 
of  the  pair  of  gears  is  the  ratio  of  the  numbers  of  teeth.  In  such 
a  case  with  a  hoist  as  shown  in  Fig.  50,  we  would  first  find  the 
mechanical  advantage  of  a  simple  windlass  with  the  crank  at- 
tached directly  to  the  drum;  then  find  the  M.  A.  of  the  pair  of 
gears,  and  by  multiplying  these  two  quantities  together  we 
would  get  the  mechanical  advantage  of  the  entire  hoist. 

PROBLEMS 


ZOO 

teoo 

.    * 

-i  _..         .                         12  ». 

FIG.  44. 


161.  The  lever  shown  in  Fig.  44  is  12  ft.  long.     Where  should  the  fulcrum 
be  placed  so  that  a  weight  of  200  lb.  will  lift  a  weight  of  1800  lb.? 


,£00 


1800 


Fia.  45. 


162.  In  Fig.  45  where  should  the  weight  of  1800  lb.  be  placed  so  that  it 
can  be  lifted  by  a  force  of  200  lb.? 


122 


SHOP  ARITHMETIC 


163.  Fig.  46  shows  a  safety  valve  V  loaded  with  a  50  Ib.  weight  at  W. 
Find  the  total  steam  pressure  on  the  bottom  of  V  necessary  to  lift  the  valve. 


F 


-f 


FIG.  46. 


164.  From  the  result  of  problem  163,  find  the  steam  pressure  per  square 
inch  if  V  is  1 J  in.  in  diameter  on  the  bottom  where  exposed  to  the  steam. 

# 


FIG.  47. 


FIG.  48. 


165.  Fig.  47  shows  the  clutch  pedal  for  an  automobile.  What  must  be 
the  length  of  the  power  arm  a  in  order  that  a  foot  pressure  of  15  Ib.  can  open 
the  clutch  against  a  spring  pressure  of  60  Ib.  having  an  arm  of  3  in.? 


5lb3. 


1000  I bs. 


Fio.  49. 


TACKLE  BLOCKS 


123 


166.  Fig.  48  shows  an  old  fashioned  windlass  for  raising  water.    If  the 
crank  is  15  in.  long,  and  the  drum  is  5  in.  in  diameter,  what  pressure  would 
be  needed  on  the  crank  to  raise  a  pail  of  water  weighing  30  lb.? 

167.  Fig.  49  represents  in  an  elementary  way  the  levers  of  a  pair  of  plat- 
form scales.     How  far  from  the  fulcrum  must  the  5  lb.  weight  be  placed  to 
balance  the  1000  lb.  weight  located  as  shown? 


FIG.  50. 


168.  The  hoist  of  Fig.  50  has  an  18  in.  crank;  the  drum  is  10  in.  in  diam- 
eter; the  diameter  of  the  large  gear  is  30  in.,  and  of  the  small  gear  6  in. 
What  weight  can  be  raised  by  a  force  of  25  lb.  on  the  crank? 


CHAPTER  XIV 
TACKLE  BLOCKS 

96.  Types  of  Blocks. — When  a  heavy  weight  is  to  be  raised  or 
moved  through  any  considerable  distance,  either  a  windlass, 
such  as  described  in  Chapter  XIII,  or  tackle  blocks  can  be  used. 
Referring  to  the  figures  in  this  chapter,  the  revolving  part  is 
called  the  Pulley  or  Sheave;  the  framework  surrounding  the 
pulleys  is  called  the  Block  and,  as  generally  used,  includes  both 
the  frame  and  the  sheaves  contained  in  it. 

In  Fig.  51  we  have  a  single  pulley  which  serves  merely  to  give 
a  change  of  direction.  There  is  no  mechanical  advantage  in  a 
single  fixed  pulley  such  as  this.  The  pull  on  the  rope  at  P  is 
transmitted  around  the  pulley  and  supports  W  on  the  other  side. 
We  can  look  at  the  pulley  in  this  case  as  a  lever  with  equal  arms. 
P  on  one  end  must  equal  W  on  the  other  end.  Such  a  block 
would  be  used  solely  for  the  convenience  it  affords,  since  it  is 
usually  easier  to  pull  down  than  up. 


124 


SHOP  ARITHMETIC 


In  Fig.  52  the  pull  P  is  in  the  same  direction  that  W  is  to  be 
moved  and  is  only  one-half  of  W.  As  explained  in  Art.  94,  the 
mechanical  advantage  of  a  machine  can  be  obtained  by  compar- 
ing the  distances  moved  by  the  force  and  the  weight.  If  a  force 
must  move  five  times  as  far  as  it  lifts  the  weight,  then  the  mechan- 
ical advantage  is  5,  and  the  force  is  one-fifth  of  the  weight.  In 
Fig.  52  the  mechanical  advantage  is  2.  This  can  be  seen  by 
raising  W  a  certain  distance.  The  rope  on  each  side  will  be 
Blacked  this  same  distances  and,  therefore,  P  must  be  drawn  up 
twice  this  distance  in  order  to  remove  the  slack.  Since  P  moves 
twice  as  far  as  W,  the  force  P  will  be  one-half  of  W,  and  the 
mechanical  advantage  will  be  2. 


I 


3 


J 


FIG.  51. 


FIG.  52. 


In  Fig.  53  we  have  merely  added,  to  the  device  of  Fig.  52,  a 
fixed  block  above  to  change  the  direction  of  P.  It  makes  no 
change  in  the  relation  of  P  and  W,  except  as  to  direction. 

In  Fig.  54  we  have  two  pulleys  in  the  fixed  block  and  two  in 
the  movable  block.  Other  cases  might  have  even  more  pulleys, 
but  the  principle  is  the  same,  and  a  general  rule  for  calculating 
their  mechanical  advantages  will  be  worked  out  for  all  cases. 
Proceeding  as  before,  let  us  imagine  that  W  and  the  movable 
block  of  Fig.  54  are  lifted  1  ft.  The  four  ropes  supporting  W 
will  each  be  slacked  1  ft.,  and  it  will  be  necessary  to  move  P  4  ft. 
to  remove  this  slack.  Hence,  the  mechanical  advantage  of  this 
system  is  4,  and  P  is  \  of  W ,  or  W  is  4  times  P. 


TACKLE  BLOCKS 


125 


In  general,  we  can  say  that  the  mechanical  advantage  is  equal 
to  the  number  of  ropes  supporting  the  movable  block  and  the 
load.  The  best  way  to  find  the  mechanical  advantage  is  to  draw 
a  sketch  of  the  blocks  and  to  count  the  number  of  ropes  that  are 
pulling  on  the  movable  block.  This  number  represents  the 
mechanical  advantage. 

Whenever  convenient,  it  is  best  to  use  as  the  movable  block 
the  one  from  which  the  free  end  of  the  rope  runs.  This  means 
that  P  will  pull  in  the  same  direction  that  W  is  to  be  moved.  The 


I 


J 


Fio.  53. 


FIG.  54. 


mechanical  advantage  is  greater  by  1  if  P  is  pulling  in  the  direc- 
tion of  motion.  Notice  in  Fig.  54  that,  if  we  turned  these  blocks 
around  and  pulled  the  other  way,  fastening  W  to  what  is  in  the 
figure  the  fixed  block,  the  mechanical  advantage  would  be  5  in- 
stead of  4. 

In  all  problems  where  there  is  any  doubt,  draw  a  rough  sketch 
and  count  the  number  of  ropes  pulling  on  the  movable  block 
(see  Fig.  55). 

Example : 

How  great  a  weight  can  be  lifted  by  a  pull  of  150  Ib.  with  a  pair  of 
pulley  blocks,  one  being  a  three  sheave  and  the  other  a  two  sheave  block? 
Calculate,  first,  using  the  three  sheave  as  the  movable  block  and,  second, 
using  the  two  sheave  block  as  the  movable  one. 


126 


SHOP  ARITHMETIC 


Explanation:  To  avoid  confusion,  the  sheaves  are  drawn  one  above  the 
other,  instead  of  parallel.  The  free  end  of  the  rope  must  run  from  the  three 
sheave  block.  Starting  from  P,  we  wind  the  rope  in  and  find  that  the  inner 
end  must  be  fastened  to  the  two  sheave  block.  We  count  the  ropes  pulling 
on  each  block  and  find  that,  with  the  three  sheave  block  as  the  movable  one, 
the  mechanical  advantage  is  6,  and  the  weight  lifted  would  be 
150X6  =  900  lb.,  First  Answer. 

With  the  two  sheave  block  as  the  movable  one,  the  mechanical  advantage 
is  5  and  the  weight  would  be 

150X5  =  750  lb.,  Second  Answer. 

In  practice,  about  60%  of  these  theoretical  weights  would  be 
raised,  the  rest  being  lost  in  overcoming  friction.  Likewise,  to 
lift  a  certain  load,  the  actual  pull  required  will  be  about  V^r  or  1  i 
of  the  theoretical  pull. 


Fia.  55. 


Fia.  56. 


97.  Differential  Pulleys. — In  lifting  heavy  weights  by  hand,  a 
very  satisfactory  apparatus  to  use  is  a  Differential  Hoist.  This 
is  a  very  simple  and  cheap  apparatus  but  it  is  not  very  efficient 
and  is,  therefore,  not  to  be  recommended  for  continuous  use. 

The  pulleys  are  arranged  as  shown  in  Fig.  56.  In  the  fixed 
block  are  two  pulleys,  A  and  B,  A  being  somewhat  larger  than  B. 


TACKLE  BLOCKS  127 

These  pulleys,  A  and  B,  are  fastened  solidly  together  and  rotate 
as  one  about  a  fixed  axis;  the  pulley  C  is  in  the  movable  block. 
An  endless  chain  passes  over  the  pulleys  as  shown,  the  rims  of 
the  pulleys  being  grooved  and  fitted  with  lugs  to  prevent  the 
chain  from  slipping.  The  loop  np  hangs  free  and  is  the  pulling 
loop. 

From  the  figure  it  is  easily  seen  that,  if  we  pull  down  on  p 
until  pulley  A  is  turned  once  around,  the  branch  m  will  be 
shortened  a  length  equal  to  the  circumference  of  A.  Since  B 
is  attached  to  A,  it  also  will  turn  once  around  and  the  branch  o 
will  be  lengthened  a  distance  equal  to  the  circumference  of  B. 

Hence,  the  loop  mo  will  be  shortened  by  an  amount  equal  to  the 
difference  of  the  circumferences  of  A  and  B;  and  the  pulley  C 
will  rise  one-half  this  amount. 

We  can  express  the  difference  in  the  distances  moved  by  m 
and  o  as 


where  D  and  d  represent  the  diameters  of  large  and  small  pulleys, 
A  and  B.  Hence  C  will  move  up  one-half  of  this  or  £  of  TT  X  (D  —  d) . 
To  cause  this  motion  of  C  upward,  the  chain  p  was  moved  a  dis- 
tance of  7TX-D. 

The  mechanical  advantage  of  the  hoist  is  obtained  by  dividing 
the  motion  of  P  by  the  motion  of  W. 

Mech.  Adv.  = 

2' 

This  can  be  simplified  by  cancelling  n  out  of  both  numerator 
and  denominator  of  the  fraction,  leaving 

Mech.  Adv.  =^ 


This  formula  might  be  written  as  a  rule  in  the  following  words: 
"  The  mechanical  advantage  of  a  differential  hoist  is  obtained  by 
dividing  the  diameter  of  the  larger  pulley  in  the  upper  block  by 
half  the  difference  between  the  diameters  of  the  larger  and 
smaller  pulleys." 

A  differential  hoist  can  actually  lift  about  30%  of  the  theoret- 
ical load  with  a  given  pull;  that  is,  the  efficiency  is  about  30%. 
Likewise,  to  lift  a  given  weight  will  require  about  -^  or  3J 


128 


SHOP  ARITHMETIC 


times  the  theoretical  force.  In  other  words,  the  actual  force 
must  be  such  that  the  30%  that  is  really  effective  will  equal  the 
theoretical  force. 

Example  : 

Calculate  the  actual  pull  required  to  lift  600  Ib.  with  a  differential 
hoist  having  10  and  8  in.  pulleys  and  an  efficiency  of  30%. 

D  =  10in. 
rf  =  8  in. 
D-rf  =  2  in. 


of  CD-d)  =  l  in. 


Mech.  Adv.  =  : 


7) 


10 


of  (D-d) 


600-7-10  =  60  Ib.  theoretical  pull 

60 -=-.30  =  60X^  =  200  Ib.  actual  pull. 
oO 

Explanation:  In  this  case  the  letters  D  and  d  of  the  formula  are  10  in. 
and  8  in.,  and  we  find  the  M.  A.  to  be  10.  It  should,  therefore,  only  require 
a  force  of  60  Ib.  to  raise  the  600  Ib.  weight.  But  we  find  that  this  type  of 
hoist  has  only  an  efficiency  of  30%,  that  is,  it  only  does  30%  of  what  we 
might  expect  it  to  do  from  pur  theories.  Then  to  lift  600  Ib.  will  require  a 
force  such  that  30%  of  it. will  be  60  Ib.  This  necessary  force  is  200  Ib. 


Of  Sheovas 


Fia.  57 


TACKLE  BLOCKS 


129 


PROBLEMS 

169.  A  weight  of  2000  Ib.  is  to  be  lifted  with  a  four  sheave  and  three 
sheave  pair  of  blocks,  as  shown  in  Fig.  57.  The  four  sheave  is  used  as  the 
movable  block.  Neglecting  friction  and  assuming  each  man  to  be  capable 
of  pulling  125  Ib.,  how  many  men  are  necessary? 


Fio.  58. 


i\  3  Sheaves 


Z  Sheaves 


Fio.  59. 

170.  A  windlass  and  tackle  blocks,  as  shown  in  Fig.  58,  are  used  for  mov- 
ing a  house.     If  the  team  can  exert  a  steady  pull  of  200  Ib.  at  the  end  of  the 
sweep,  find  the  theoretical  pull  on  the  house.     Also  find  the  actual  pull  on 
the  house  if  the  efficiency  of  the  whole  mechanism  is  65%. 

171.  Draw  a  sketch  of  a  pair  of  blocks,  each  having  3  pulleys,  and  indicate 
which  should  be  the  movable  block  in  order  to  secure  the  greatest  mechanical 
advantage.     What  would  the  mechanical  advantage  be? 


130  SHOP  ARITHMETIC 

172.  When  the  geared  windlass  of  the  dimensions  shown  in  Fig.  59  is 
used  with  the  pair  of  pulley  blocks,  find  the  weight  at  W  that  can  be  lifted 
by  a  force  of  25  Ib.  on  the  crank. 

173.  A  differential  hoist  has  pulleys  1\  in.  and  6  in.  in  diameter.     We 
attach  a  weight  of  200  Ib.  to  the  hoist  and  find  that  a  pull  of  58  Ib.  is  re- 
quired to  raise  the  weight. 

(a)  Find  the  theoretical  force  required  to  raise  200  Ib.  with 
this  hoist. 

(b)  From  this  and  the  force  actually  required,  calculate  the 
efficiency  of  the  hoist. 

174.  Three  men  pull  70  Ib.  apiece  on  a  pair  of  pulley  blocks,  two  sheaves 
above  and  one  below.     The  single  block  is  movable.    Find  the  weight  that 
can  be  lifted: 

(a)  Neglecting  friction; 

(b)  Assuming  that  40%  of  the  work  is  lost  in  friction. 

175.  A  load  of  2  tons  is  to  be  lifted  with  a  differential  hoist.     The  pulleys 
are  12  in.  and  10J  in.  in  diameter. 

(a)  What  is  the  theoretical  pull  required  to  lift  the  load? 

(b)  What  is  the  actual  pull  required,  if  the  efficiency  of  the 
hoist  is  30%? 


CHAPTER  XV 
THE  INCLINED  PLANE  AND  SCREW 

98.  The  Use  of  Inclined  Planes. — An  Inclined  Plane  is  a  surface 
which  slopes  or  is  inclined  from  the  horizontal.     Any  one  who 
has  had  experience  in  raising  heavy  bodies  from  one  level  to 
another  knows  that  inclined  planes  are  very  useful  for  such  work. 
The  Wedge  is  a  form  of  an  inclined  plane,  the  powerful  effect  of 
which  in  splitting  wood,  quarrying  stone,  aligning  machinery, 
and  performing  many  other  heavy  duties  is  well  known.     The 
inclined  plane,  like  the  lever  and  the  tackle  block,  enables  us  to 
lift  a  heavy  weight  with  a  smaller  force. 

99.  Theory  of  the  Inclined  Plane. — The  work  done  in  moving  a 
body  up  an  inclined  plane  is  merely  the  work  of  raising  the  body 
vertically.     If  we  skid  an  engine  base  from  the  shop  floor  onto  a 
flat  car,  the  work  accomplished  is  the  raising  of  the  base  from 
the  floor  level  to  the  car  level,  and  is  the  same  as  if  it  was  raised 
straight  up  by  a  crane,  or  by  tackle  blocks.     The  effect  of  the 
long  incline  is  similar  to  that  of  a  long  force  arm  on  a  lever.     It 
enables  the  force  doing  the  work  to  use  a  greater  distance,  and 
hence  the  force  will  be  smaller  than  the  weight  raised. 

Neglecting  friction  or,  in  other  words,  supposing  bodies  to  be 
perfectly  smooth  and  hard,  no  work  is  done  in  moving  the  bodies 


THE  INCLINED  PLANE  AND  SCREW 


131 


in  a  horizontal  direction;  hence  the  work  done  upon  a  body  when 
it  is  moved  equals  the  weight  of  the  body  times  the  vertical 
height  to  which  it  is  raised.  In  studying  the  theory  of  the 
inclined  plane,  we  find  that  the  force  generally  acts  in  one  of  two 
directions  in  raising  the  body:  either  parallel  to  the  incline  or 
parallel  to  the  horizontal  base. 

In  Case  I  (Fig.  60)  the  force  P  is  exerted  along  the  incline,  and, 
in  raising  the  weight  to  the  top,  will  act  through  a  distance  I. 
Meamvhile,  it  will  raise  W  a  distance  h.  Consequently,  the 

I 

mechanical  advantage  will  be  -• 

h 

If  we  remember  that  the  work  put  in  equals  the  work  got  out 
of  a  machine  (neglecting  what  is  lost  in  friction)  we  see  that  we 
have  the  formula 


To  sum  up,  when  the  force  is  exerted  parallel  to  the  surface  of 
the  inclined  plane,  the  force  times  the  length  of  the  inclined 
plane  equals  the  weight  times  the  vertical  height  through  which 


I     F~OftO£     f     t-MTAt.t.E.i.    TO 


Fio.  60. 


TOTHX:  BASE.. 
FIG.  61. 


the  weight  is  raised  by  the  plane;  or  the  mechanical  advantage 
equals  the  length  of  the  inclined  surface  divided  by  the  height. 
If  the  weight  to  be  raised  is  great  as  compared  with  the  force 
available,  a  comparatively  long  incline  must  be  used  to  give  the 
necessary  mechanical  advantage. 

In  Case  II  (Fig.  61)  the  force  acts  parallel  to  the  base  of  the 
inclined  plane;  that  is,  along  the  horizontal.  This  case  is  not 
often  found  in  this  elementary  form,  but  is  seen  in  jack  screws,  in 
worm  gearing,  in  wedges,  and  in  cams,  all  of  which  are  modifica- 
tions of  inclined  planes.  When  the  force  acts  parallel  to  the 
base,  the  work  expended  by  it  is  the  product  of  the  force  times 
the  length  of  the  base;  the  work  accomplished  is,  as  before,  the 
product  of  the  weight  times  the  height. 


11 


Mechanical  advantage  =  -r 


132  SHOP  ARITHMETIC 

A  comparison  of  these  formulas  with  those  for  Case  I  shows 
that  the  mechanical  advantage  is  greatest  where  the  force  P  is 
exerted  along  the  incline,  as  in  Case  I,  because  I,  the  length  of  the 
incline  or  the  hypotenuse  of  a  right  triangle,  is  greater  than  6, 
the  length  of  the  base. 

There  may  be  other  cases  where  the  force  acts  in  some  other 
direction,  but  they  are  seldom  seen  in  practice. 

100.  The  Wedge. — The  Wedge  consists  of  two  inclined  planes 
placed  base  to  base,  the  force  acting  parallel  to  the  base,  as 
shown  in  Fig.  62,  where  the  horizontal  center  line  of  the  wedge  is 


FIG.  62. 

the  common  base  of  the  two  inclined  planes.  Usually  the  wedge 
is  moved  instead  of  the  object  to  be  raised,  but  the  effect  is  the 
same  and  the  force  relations  are  the  same  as  if  the  object  itself 
were  being  moved  up  a  stationary  incline.  In  Fig.  62  it  will  be 
seen  that  the  weight  W  will  be  raised  a  distance  h  when  the 
wedge  is  driven  a  distance  I.  The  work  expended  in  driving  the 
wedge  is  PXl;  the  work  accomplished  in  raising  the  weight  is 
WXh;  and,  neglecting  friction,  these  are  equal,  or 


From  this  we  see  that  the  mechanical  advantage  of  the  wedge  is  : 
Mechanical  advantage  =  j- 

The  relation  of  P  and  W  might,  if  desired,  be  written  as  a  pro- 
portion, as  follows: 

P:W  =  h:l 
Example  : 

Fig.  63  shows  an  adjustable  pillow  block  for  a  Corliss  engine,  the 
bearing  being  raised  or  lowered  by  means  of  the  wedge  underneath.  If  the 
weight  of  the  shaft  and  the  fly-wheel  upon  this  bearing  is  6000  lb.,  and  the 
wedge  has  a'taper  of  1  in.  per  foot  of  length,  what  pressure  must  be  exerted 
on  the  wedge  by  the  screw  S  in  raising  the  shaft? 

Mech.  Adv.  =  12 

6000  -4-12     =500  lb.  Answer. 


THE  INCLINED  PLANE  AND  SCREW 


133 


Explanation:  If  the  taper  of  the  wedge  is  1  in.  in  12  in.,  then  a  motion 
of  1  ft.  would  raise  the  bearing  1  in.;  or  a  motion  of  1  in.  in  the  wedge  would 
raise  the  bearing  fa  in.  Hence,  the  Mech.  Adv.  of  the  wedge  is  12  and 

W 
p=  ^  =  500  Ib. 

\2t 

Note. — There  would  also  be  required,  in  addition  to  this  500  lb.,  a  force 
sufficient  to  overcome  the  friction  on  the  top  and  bottom  of  the  wedge, 
which  is  neglected  in  this  solution. 


FIG.  63. 

101.  The  Jack  Screw. — A  screw  is  nothing  but  an  inclined 
plane  which,  instead  of  being  straight,  is  wrapped  around  or  cut 
into  a  round  rod  or  bar.  Turning  the  screw  gives  the  same 
effect  as  giving  a  straight  push  on  an  inclined  plane  or  wedge. 

When  we  raise  an  object  with  a  jack  screw,  such  as  shown  in 
Fig.  64,  the  weight  presses  down  on  the  screw  and,  consequently, 

I////////////J '///////////, 
P  <- — -> 


Fia.  84. 

is  borne  by  the  threads  (which  are  the  inclined  planes).  The 
threads  are  advanced  and  the  weight  is  raised  by  a  pull  (which 
we  will  call  P)  on  the  end  of  the  rod  whose  length  is  marked  R. 
The  distance  the  weight  W  is  moved  for  one  revolution  of  the 
screw  equals  the  lead  of  the  screw  expressed  as  a  fraction  of  an 
Inch.  The  lead  of  a  screw  is  the  distance  it  advances  lengthwise 
in  one  turn  or  revolution.  The  force  P  moves  through  a  distance 


134  SHOP  ARITHMETIC 

equal  to  the  circumference  of  a  circle  whose  radius  is  the  length 
of  the  handle;  if  we  represent  the  length  of  the  handle  or  lever  by 
R,  then  the  distance  traversed  by  P  in  one  revolution  is  TT  X  2  X  R. 
If  we  let  the  letter  L  represent  the  lead  of  the  screw  then  we  will 
have  the  work  accomplished  in  one  revolution  of  the  screw  = 
WxL.  Meanwhile,  the  work  expended  in  doing  it 

=  PXKX2XR. 
Assuming  that  there  is  no  friction  in  the  screw,  we  have 


Distance  P  moves         7TX2X.R 
or    Mech.  Adv.  =^—  -  ^r~-  --  =  —  r=  - 

Distance  W  is  raised  L 

If  stated  in  words,  these  formulas  would  read:  "The  force 
multiplied  by  the  circumference  of  the  circle  through  which  it 
moves  equals  the  weight  multiplied  by  the  lead  of  the  screw." 

"The  mechanical  advantage  of  a  jack  screw  equals  the  cir- 
cumference of  the  circle  through  which  the  force  moves  divided 
by  the  lead  of  the  screw"  (the  amount  the  screw  advances  in  one 
turn)  . 

Example  : 

With  a  1$  in.  jack  screw  having  3  threads  per  inch  and  a  pull  of 
50  Ib.  at  a  radius  of  18  in.,  calculate: 

(a)  The  theoretical  load  that  can  be  lifted  by  the  screw; 

(6)   The  actual  load  if  the  efficiency  of  the  screw  is  18%. 


(a)  Mech.  Adv.  = 
.R  =  18  in.  and 


L 

1  . 
=  3  in. 

Hence, 

3.1416X2X18 


,.    ,      .  , 
Mech.  Adv.  = 


= 

3 
_ 

—  OOJ7 


_! 

3 

W  =  339X50  =  16050  Ib.,  Answer. 
(b)   18%  of  16950  =  3051  Ib.,  Answer. 

Explanation:  If  there  are  3  threads  per  inch,  the  lead  is  $  in.,  and  if  the 
radius  is  18  in.,  we  have  the  Mech.  Adv.  =339.3.  In  theory  then  we  should 
be  able  to  lift  50  Ib.  X  339  =  16950  Ib.  with  this  screw.  But  a  screw  has  con- 
siderable friction  and  for  this  reason  only  18%  of  the  energy  expended  in 
this  case  is  effective,  the  remaining  82%  being  all  lost  in  friction.  The 
actual  weight  lifted  is,  therefore,  only  18%  of  16950  Ib.  or  3051  Ib. 

102.  Efficiencies.  —  In  explaining  the  machines  of  this  chapter 
and  of  Chapters  XIII  and  XIV,  it  was  assumed  that  no  work  is 
lost  in  friction  within  the  machines.  In  a  properly  mounted  lever 
there  is  little  energy  lost.  In  a  tackle  block  the  loss  depends  on 


THE  INCLINED  PLANE  AND  SCREW  135 

the  size  of  the  pulleys  as  compared  with  the  size  of  the  rope  and 
on  the  nature  of  the  pulley  bearings.  The  efficiency  may  vary 
from  60  to  95  %.  The  more  pulleys  there  are,  the  lower  will  be 
the  efficiency,  because  each  bend  in  the  rope  and  each  pulley 
means  a  loss  in  friction. 

With  inclined  planes,  the  efficiency  may  vary  all  the  way 
from  0  to  nearly  100%.  It  will  be  lowest  if  the  weight  is  merely 
slid  on  the  plane  and  will  be  much  higher  if  wheels  or  rollers  are* 
used. 

In  any  machine,  if  the  weight  will  start  back  of  its  own  accord 
when  the  force  is  removed,  the  friction  is  less  than  50%  and  the 
efficiency  is  greater  than  50%.  If  the  weight  will  not  start  back, 
the  efficiency  is  less  than  50%.  This  can  be  shown  as  follows: 
Of  the  force  applied  to  a  machine,  part  of  it  is  absorbed  in  over- 
coming the  friction  within  the  machine.  The  balance  goes 
through  the  machine  and  is  effective  in  accomplishing  the  work 
to  be  done.  Of  the  whole  force  applied,  the  per  cent  which  this 
effective  force  represents  is  called  the  Efficiency.  If  the  efficiency 
is  less  than  50%,  it  shows  that  the  friction  absorbs  more  than 
half  of  the  total  force  and,  therefore,  that  the  friction  is  greater 
than  the  effective  force.  Now,  suppose  we  had  a  simple  machine 
such  as  a  jack-screw,  being  used  to  raise  a  weight.  If  the  applied 
force  is  removed,  the  friction  will  remain  the  same,  but  will  now 
act  to  hold  the  weight  from  'going  back.  If  the  friction  is  suffi- 
cient to  hold  the  weight,  it  must  at  least  equal  the  effective  or 
theoretical  force  required  to  raise  the  weight.  Therefore,  if  a 
machine  does  not  run  backward  when  the  force  is  removed,  the 
friction  must  be  more  than  one-half  of  the  total  force  required  to 
raise  the  weight,  and  the  effective  force  must  be  less  than  one- 
half  of  this  total  force.  Hence,  the  efficiency  in  such  a  case  is 
less  than  50%.  A  jack-screw  will  not  go  down  of  its  own  accord 
when  the  force  is  removed  and  therefore  its  efficiency  is  less  than 
50%.  In  reality,  for  the  usual  dimensions  of  screws,  it  has  been 
found  to  be  only  from  15  to  20%.  Mr.  Wilfred  Lewis  has 
derived,  from  experiment,  a  simple  formula  which  gives  the 
average  efficiency  for  a  jack-screw  under  ordinary  conditions. 

L 


in  which  E  is  the  efficiency,  as  a  decimal, 
where  L  is  the  lead  of  the  screw, 
and  D  is  the  diameter  of  the  screw. 
12 


136 


SHOP  ARITHMETIC 


Example : 

Find  the  probable  efficiency  of  the  screw  given  in  the  example 
under  Article  101. 

T      l  •  A  n      i1- 

jL/=77  in.,  and  D  =  IH  in. 
o  ~ 

.33 


.33  +  1.5 

.#  =  18%,  Answer. 

One  can  get  an  approximate  idea  of  the  efficiency  of  any 
machine  by  observing,  as  before  explained,  whether  or  not  it  will 
run  backward  of  its  own  accord  when  the  force  is  removed. 
This  will  tell  whether  the  efficiency  is  above  or  below  50%.  If 
it  is  above  50%  and  a  considerable  force  is  required  to  keep  the 
weight  from  going  back,  then  the  efficiency  is  high.  If,  however, 
a  very  slight  pull  will  hold  it  from  going  back,  then  the  efficiency 
is  not  very  much  above  50%.  If  we  find  the  efficiency  to  be 
under  50%  but  find  that  only  a  very  small  pull  will  start  the 
weight  down,  then  the  efficiency  is  not  far  under  50%.  On  the 
other  hand,  if  it  seems  as  if  almost  as  great  a  force  is  required  to 
lower  the  weight  as  to  raise  it,  this  signifies  that  the  efficiency  of 
the  machine  is  extremely  low. 

PROBLEMS 

176.  An  engine  weighing  5  tons  is  to  be  loaded  onto  a  car,  the  floor  of 
which  is  6  ft.  from  the  ground.     If  16  ft.  timbers  are  used  for  the  runway, 
find  the  pull  necessary  to  draw  the  engine  up  the  slope,  neglecting  friction. 

177.  How  many  pounds  must  a  locomotive  exert  to  pull  a  train  of  50 
cars,  each  weighing  50  tons,  up  a  grade  of  3  in.  in  100  ft.? 

178.  A  building  is  to  be  raised  by  means  of  4  jack-screws;  the  screws  are 
2  in.  in  diameter,  with  4  threads  to  the  inch.     The  lever  is  20  in.  long  and  a 
30  Ib.  force  is  required  on  each  handle.     Calculate  th6  theoretical  weight 
which  the  four  screws  should  lift  under  these  conditions. 

179.  Calculate  the  probable  efficiency  of  these  jack-screws  from  Lewis' 
formula  and  estimate  the  probable  weight  of  the  building. 


FIG.  65. 


180.  A  windlass  with  an  axle  8  in.  in  diameter  and  crank  18  in.  long  is 
used  in  connection  with  an  inclined  plane  20  ft.  long  and  5  ft.  high,  as  shown 
in  Fig.  65.  Neglecting  friction,  what  weight  can  be  pulled  up  the  slope 
with  a  force  of  150  Ib.  on  the  crank? 


CHAPTER  XVI 
WORK,  POWER,  AND  ENERGY;  HORSE-POWER  OF  BELTING 

103.  Work. — Whenever  a  force  causes  a  body  to  move,  work 
is  done.     Unless  the  body  is  moved,  no  work  is  accomplished. 
A  man  may  push  against  a  heavy  casting  for  hours  and,  unless 
he  moves  it,  he  does  no  work,  no  matter  how  tired  he  may  feel  at 
the  end  of  the  time.     It  is  evident  that  there  are  two  factors  to 
be    considered    in    measuring    work — force    and    distance.     In 
the  study  of  levers,  tackle  blocks,  and  inclined  planes  we  dealt 
with  the  problem  of  work.     In  any  of  these  machines  the  work 
accomplished  in  lifting  a  weight  is  measured  by  the  product  of 
the  weight  and  the  distance  it  is  moved.     The  work  expended 
or  put  into  the  machine  to  accomplish  this  is  the  product  of  the 
force  exerted  times  the  distance  through  which  this  force  must 
act.     We  found  that,  if  we  neglect  the  work  lost  in  friction,  the 
work  put  into  a  machine  is  equal  to  the  work  accomplished  by  it. 
The  actual  difference  between  the  work  put  in  and  the  work 
accomplished  is  the  amount  that  is  lost  in  friction.     The  follow- 
ing expressions  may  make  these  relations  clearer: 

Work  lost  in  Friction  =  Work  put  in  — Work  got  out. 

_~  .  Work  got  out 

Efficiency  =  ^ — r-&  — — 
Work  put  in 

104.  Unit  of  Work. — The  unit  by  which  work  is  measured  is 
called  the  Foot-pound.     This  is  the  work  done  in  overcoming  a 
resistance  of  one  pound  through  a  distance  of  1  ft.;  that  is,  if  a 
weight  of  1  Ib.  is  lifted  1  ft.,  the  work  done  is  equal  to  1  foot- 
pound.    All  work  is  measured  by  this  standard.     The  work  in 
foot-pounds  is  the  product  of  the  force  in  pounds  and  the  distance 
in  feet  through  which  it  acts.     In  lifting  a  weight  vertically,  the 
resistance,  and  hence,  the  force  that  must  be  exerted,  is  equal  to 
the  weight  itself  in  pounds.     The  work  done  is  the  product  of 
the  weight  times  the  vertical  distance  that  it  is  raised.     If  a 
weight  of  80  Ib.  is  lifted  a  distance  of  4  ft.,  the  work  done  is  80X4 
or  320  foot-pounds.     It  would  require  this  same  amount  of  work 
to  lift  40  Ib.  8  ft.,  or  to  lift  20  Ib.  16  ft. 

13  137 


138  SHOP  ARITHMETIC 

When  a  body  s  moved  horizontally,  the  only  resistance  to  be 
overcome  is  the  friction.  When  a  team  of  horses  pulls  a  loaded 
wagon,  the  only  resistances  which  it  must  overcome  are  the 
friction  between  the  wheels  and  the  axles,  and  the  resistance  on 
the  tires  caused  by  the  unevenness  of  the  road. 

The  work  necessary  to  pump  a  certain  amount  of  water  is  the 
weight  of  the  water  times  the  height  through  which  it  is  lifted  or 
pumped  (plus,  of  course,  the  work  lost  in  friction  in  the  pipes). 
The  work  necessary  to  hoist  a  casting  is  the  weight  of  the  casting 
times  the  height  to  which  it  is  lifted.  The  work  done  by  a  belt 
is  the  effective  pull  of  the  belt  times  the  distance  in  feet  which 
the  belt  travels.  The  work  done  in  hoisting  an  elevator  is  the 
weight  of  the  cage  and  of  the  load  it  carries  times  the  height  of 
the  lift.  Numerous  other  illustrations  of  work  will  suggest 
themselves  to  the  student. 

105.  Power. — Power  is  the  rate  of  doing  work;  that  is,  in  cal- 
culating power  the  time  required  to  do  a  certain  number  of  foot- 
pounds of  work  is  considered.  If  10,000  Ib.  are  lifted  7  ft.  the 
work  done  is  70,000  foot-pounds,  regardless  of  how  long  it  takes. 
But,  if  one  of  two  machines  can  do  this  in  one-half  the  time  that 
the  other  machine  requires,  then  the  first  machine  has  twice  the 
power  of  the  second. 

The  engineer's  standard  of  power  is  the  Horse-power,  which  may 
be  defined  as  the  ability  to  do  33, 000  foot-pounds  of  work  per  minute. 
The  horse-power  required  to  perform  a  certain  amount  of  work 
is  found  by  dividing  the  foot-pounds  done  per  minute  by  33,000. 
If  an  engine  can  do  1,980,000  foot-pounds  in  a  minute,  its  horse- 
power would  be  1,980,000^-33,000  =  60.  An  engine  that  can 
raise  66,000  Ib.  to  a  height  of  10  ft.  in  1  minute  will  do  66,000  Ib. 
XlO  ft.  =660,000  foot-pounds  per  minute,  and  this  will  equal 
af§f$£=20  horse-power.  If  another  engine  takes  4  minutes  to 
do  this  same  amount  of  work,  it  is  only  one-fourth  as  powerful; 
the  work  done  per  minute  will  be  &&S>£AQ.  =  165,000  foot-pounds 
per  minute;  and  its  horse-power  is  -LH7nnj-  =  5  horse-power. 

Example : 

An  electric  crane  lifts  a  casting  weighing  3  tons  to  a  height  of  20  ft. 
from  the  floor  in  30  seconds;  what  is  the  horse-power  used? 

3  tons  =  6000  Ibs. 

6000  Ib.  X20  ft.  =  120,000  foot-pounds  done. 
120,000  foot-pounds  done  in  30  seconds  = 
240,000  foot-pounds  per  1  minute. 

=  7.27  horse-power  used. 


WORK,  POWER,  AND  ENERGY  139 

106.  Horse -power  of  Belting. — A  belt  is  an  apparatus  for  the 
transmission  of  power  from  one  shaft  to  another.  The  driving 
pulley  exerts  a  certain  pull  in  the  belt  and  this  pull  is  transmitted 
by  the  belt  and  exerted  on  the  rim  of  the  driven  pulley. 

The  power  transmitted  by  any  belt  depends  on  two  things — 
the  effective  pull  of  the  belt  tending  to  turn  the  wheel,  and  the 
speed  with  which  the  belt  travels.  From  the  preceding  pages,  it 
is  easily  seen  that  these  include  the  three  items  necessary  to 
measure  power.  The  pull  of  the  belt  is  the  force.  The  speed, 
given  in  feet  per  minute,  includes  both  distance  and  time.  Force, 
distance  and  time  are  the  three  items  necessary  for  the  measure- 
ment of  power. 

The  total  pull  that  a  belt  will  stand  depends  on  its  width  and 
thickness.  It  should  be  wide  enough  and  heavy  enough  to 
stand  for  a  reasonable  time  the  greatest  tension  put  upon  it. 
This  is,  of  course,  the  tension  on  the  driving  side.  This  tension, 
however,  does  not  represent  the  force  tending  to  turn  the  pulley. 
The  force  tending  to  turn  the  pulley  (or  the  Effective  Pull,  as  it 
is  called)  is. the  difference  in  tension  between  the  tight  and  the 
slack  sides  of  the  belt. 

The  effective  pull  that  can  be  allowed  in  a  belt  depends  prima- 
rily on  the  width,  thickness,  and  strength  of  the  leather,  or  what- 
ever material  the  belt  is  made  of.  Besides,  we  must  consider 
that  every  time  a  belt  causes  trouble  from  breaking  or  becoming 
loose,  it  means  a  considerable  loss  in  time  of  the  machine,  of 
the  men  who  are  using  it,  and  of  the  men  required  to  make  the 
repairs  and,  therefore,  it  should  not  be  loaded  as  heavily  as 
might  otherwise  be  allowed.  Leather  belts  are  called  "single/' 
"double,"  "triple,"  or  "quadruple,"  according  to  whether  they 
are  made  of  one,  two,  three,  or  four  thicknesses  of  leather. 
Good  practice  allows  an  effective  pull  of  35  Ib.  in  a  single  leather 
belt  per  inch  of  width.  In  a  double  belt  a  pull  of  70  Ib.  per  inch 
of  width  may  be  allowed.  The  pull  times  the  width  gives  the 
total  effective  pull  or  the  force  transmitted  by  the  belt. 

The  force  times  the  velocity,  or  speed,  of  the  belt  in  feet  per 
minute  will  give  the  foot-pounds  transmitted  by  it  in  1  minute. 
One  horse-power  is  a  rate  of  33,000  foot-pounds  per  minute; 
hence,  the  horse-power  of  a  belt  is  obtained  by  dividing  the 
foot-pounds  transmitted  by  it  per  minute  by  33,000.  The 
velocity  of  the  belt  is  calculated  from  the  diameter  and  revolu- 
tions per  minute  of  either  one  of  the  pulleys  over  which  the  belt 


140  SHOP  ARITHMETIC 

travels,  as  explained  in  Chapter  VII.  From  these  considerations, 
the  formula  for  the  horse-power  that  a  belt  will  transmit  may 
be  written 

PXWXV 


H  = 


33000 


where  H  =  horse-power 

P  =  effective  pull  allowed  per  inch  of  width 

W  =  width  in  inches 

V  =  velocity  in  feet  per  minute 

Stated  in  words,  this  formula  would  read  as  follows:  "The  horse- 
power that  may  be  transmitted  by  a  belt  is  found  by  multiplying 
together  the  allowable  pull  per  inch  of  width  of  the  belt,  the 
width  of  the  belt  in  inches,  and  the  velocity  of  the  belt  in  feet  per 
minute  and  then  dividing  this  product  by  33,000. 

Example  : 

Find  the  horse-power  that  should  be  carried  by  a  12-in.  double 
leather  belt,  if  one  of  the  pulleys  is  14  in.  in  diameter  and  runs  1100  R.  P.  M. 

Explanation:    To  get  the  horse- 

P  =  70  Ib.  power,  we  must  first  find  the  values 

W  =  12  in.  Of  Pt  W>  and  F.     We  will  take  P 

v  —  »,N/^  vunn  as  70  Ib.  since  this  is  a  double  belt. 

r    —  71  A  Vr»  •*•  •*•  AUU  Tr,  .        .  ,  „  .  T,     .-,  |       ., 

12  W  is  given,  12  in.    V,  the  velocity, 

=  4032  ft.  per  min.  is  obtained  by  multiplying  the  cir- 

TT_PXWX  V  cumference  of  the  pulley  by  the  R. 

33000  P-  M.,  which  gives  us  4032.    Multi- 

70X12X4032  plying  these  three  together  gives 

=  -  oonnn  3,386,880  foot-pounds  per  minute, 

-102+  horse-Dower  Answer          and  dividing  by  33,000  we  have 
>e-power,  Answer.         102  +  as  the  horse-power  that  this 

belt  might  be  required  to  carry. 

107.  Widths  of  Belts.  —  It  is  possible,  also,  to  develop  a 
formula  with  which  to  calculate  the  width  of  belt  required  to 
transmit  a  certain  horse-power  at  a  given  velocity. 

One  horse-power  is  33,000  foot-pounds  per  minute.  Then  the 
given  number  of  horse-power  multiplied  by  33,000  gives  the 
number  of  foot-pounds  to  be  transmitted  per  minute. 

Foot-pounds  per  minute  =  33000  X  H 

If  we  know  the  velocity  in  feet  per  minute,  we  can  divide  the 
foot-pounds  per  minute  by  the  velocity;  the  quotient  will  be  the 
force  or  the  effective  pull  in  the  belt. 


Force  ^ 

X      \J±   V/V>     *—  y- 


WORK,  POWER,  AND  ENERGY  141 

Now  the  force  can  be  divided  by  the  allowable  pull  per  inch  of 
width  of  belt.     The  result  will  be  the  necessary  width. 

,„    33000  XH 


Stated  in  words,  this  formula  would  read:  "To  obtain  the  width 
of  belt  necessary  for  a  certain  horse-power;  multiply  the  horse- 
power by  33,000  and  divide  by  the  product  of  the  allowable  pull 
per  inch  of  width  of  belt  times  the  velocity  of  the  belt  in  feet  per 
minute." 

Example  : 

Find  the  width  of  a  single  belt  to  transmit  10  horse-power  at  a 
speed  of  2000  ft.  per  minute. 

II  =  10  Explanation:    We  have  given  the  horse- 

V  =  2000  power  and  the  velocity,  and  we  know  that 

P  =  35  for  a  single  belt  a  pull  of  35  Ib.  per  inch  is  al- 

33000  XH  lowable.     This  data  is  all  that  is  needed  to 

PXV  calculate  the  width,  which  comes  out  4£  in. 
The  next  larger  standard  width  is  5  in.,  so 

2  that  is  the  size  that  would  be  used. 

33000  X^  33          . 

~  ??X#W>  =  7 

7 
Use  5  in.  belt,  Answer. 

108.  Rules  for  Belting.  1.  Belt  Thickness.  —  It  is  generally 
advisable  to  use  single  belting  in  all  cases  where  one  or  both 
pulleys  are  under  12  in.  in  diameter,  and  double  belting  on  pulleys 
12  in.  or  larger.  Triple  and  quadruple  belts  are  used  only  for 
main  drives  where  considerable  power  is  to  be  transmitted  and 
where  a  single  or  double  belt  would  have  an  excessive  width.  A 
triple  belt  should  not  be  run  on  a  pulley  less  than  20  in.  in  diam- 
eter, nor  a  quadruple  belt  on  a  pulley  less  than  30  in.  in  diameter. 

2.  Tension  per  Inch  of  Width.  —  An  effective  pull  of  35  Ib.  per 
inch  of  width  of  belt  is  allowable  for  single  belts.     For  double 
belts  an  effective  pull  of  70  Ib.  per  inch  is  allowable  unless  the 
belt  is  used  over  a  pulley  less  than  12  in.  in  diameter,  in  which 
case  only  50  Ib.  per  inch  should  be  allowed.     A  prominent  manu- 
facturer of  rubber  belting  recommends  33  Ib.  per  inch  of  width  of 
belt  for  4-ply  belts  and  43  Ib.  for  6-ply  rubber  belts. 

3.  Belt  Speeds.  —  The  most  efficient  speed  for  belts  to  run  is 
from  4000  to  4500  ft.  per  minute.     Belts  will  not  hug  the  pulley 
and  therefore  will  slip  badly  if  run  at  a  speed  of  over  one  mile 
per  minute.     These  figures  are  seldom  reached  in  machine  shops. 


142 


SHOP  ARITHMETIC 


Belts  for  machine  tool  drives  run  from  1000  to  2000  ft.  per  min- 
ute, while  main  driving  belts  for  line  shafts  are  more  often  run 
about  3000  ft.  per  minute.  On  wood-working  tools  we  find 
higher  speeds,  usually  4000  ft.  per  minute  or  over. 

4.  Distance  between  Centers. — The  best  distance  to  have  be- 
tween the  centers  of  shafts  to  be  connected  by  belting  is  20  to  25 
ft.     For  narrow  belts  and  small  pulleys  this  distance  should  be 
reduced. 

5.  Arrangement  of  Pulleys. — It  is  desirable  that  the  angle  of 
the  belt  with  the  floor  should  not  exceed  45  degrees;  that  is,  the 
belt  should  be  nearer  horizontal  than  vertical.     Fig.  66  shows 


Driver 


Fio.  67. 


the  effect  of  having  a  belt  nearly  vertical.  Any  sag  in  the  belt 
causes  it  to  drop  away  from  the  lower  pulley  and  lose  its  grip  on 
it.  Fig.  67  shows  the  best  arrangement.  Have  the  belt  some- 
where near  horizontal  and  have  the  tight  side  of  the  belt  under- 
neath, if  possible.  This  will  increase  the  wrap  of  the  belt  around 
the  pulleys.  If  the  lower  side  is  the  loose  side,  the  wrap  will  be 
decreased  by  the  sag. 

It  is  also  desirable,  whenever  possible,  to  arrange  the  shafting 
and  machinery  so  that  the  belts  will  run  in  opposite  directions 
from  the  shaft,  as  shown  in  Fig.  68.  This  arrangement  balances 
somewhat  the  belt  pulls,  and  reduces  the  friction  and  wear  in 
the  bearings. 


WORK,  POWER,  AND  ENERGY 


143 


For  belts  which  are  to  be  shifted,  the  pulley  faces  should  be 
flat;  all  other  pulleys  should  have  the  faces  crowned  (high  in  the 
center)  about  T8¥  in.  per  foot  of  width. 


I.  ! 


FIG.  68. 

6.  Grain  and  Flesh  Sides. — The  grain  side  of  the  leather  is  the 
side  from  which  the  hair  is  removed.  It  is  the  smoothest  but 
weakest  side  of  the  leather,  and  should  run  next  to  the  pulley 
surface.  It  will  wrap  closer  to  the  pulley  surface  and  thus  get  a 
better  grip  on  the  pulley.  Furthermore,  the  flesh  side,  being 
stronger,  is  better  able  to  stand  the  stretching  which  must  occur  in 
the  outside  of  the  belt  in  bending  around  a  pulley. 


oursioc. 


Fio.  69. 


7.  Belt  Joints. — Whenever  possible,  the  ends  of  belts  should 
be  fastened  together  by  splicing  and  cementing.  Never  run  a 
wide  cemented  belt  onto  the  pulleys  as  one  side  is  liable  to  be 
stretched  out  of  true.  Rather  lift  one  shaft  out  of  the  bearings, 


144  SHOP  ARITHMETIC 

place  the  belt  on  the  pulleys,  and  force  the  shaft  back  into  place. 
Of  other  methods  of  fastening  belts,  the  leather  lacing  is  un- 
doubtedly the  best  when  properly  done.  In  lacing  a  belt,  begin 
at  the  center  and  lace  both  ways  with  equal  tension.  Fig.  69 
shows  an  excellent  method  of  lacing  belts.  The  lacing  should 
be  crossed  on  the  outside  of  the  belt.  On  the  inside,  the  lacing 
should  lie  in  line  with  the  belt.  Holes  should  be  about  1£  in. 
apart  and  their  edges  should  be  at  least  $•  in.  from  the  ends  of 
the  belt.  The  holes  should  be  punched,  preferably  with  an 
oval  punch,  the  long  dimension  of  the  oval  running  lengthwise  of 
the  belt  so  as  not  to  weaken  the  belt  too  much. 

PROBLEMS 

181.  A  casting  weighs  300  Ib.     How  much  work  is  required  to  place  it  on 
a  planer  bed  3  ft.  5  in.  above  the  floor? 

182.  How  much  work  is  required  to  pump  5000  gallons  of  water  into  a 
tank  150  ft.  above  the  pump? 

183.  Find  the  horse-power  that  may  be  transmitted  per  inch  of  width  by 
a  single  belt  running  at  2500  ft.  per  minute.     How  does  this  compare  with 
a  double  belt  running  at  the  same  speed? 

184.  A  6  in.  double  belt  is  carried  by  a  48  in.  pulley  running  250  R.  P.  M. 
Find  the  horse-power  that  may  be  transmitted. 

185.  A  shop  requires  50  horse-power  to  run  it.     The  main  shaft  runs 
250  R.  P.  M.     Select  a  main  driving  pulley  and  determine  width  of  double 
belt  to  run  the  shop. 

186.  A  foundry  fan  runs  3145  R.  P.  M.,  and  requires  24  horse-power  to 
run  it.     There  are  two  single  belts  on  the  blower  running  over  pulleys  7  in. 
in  diameter.     Determine  the  necessary  width  of  belt. 

Note. — (Each  belt  should  be  wide  enough  to  drive  the  fan  so  that  in  case 
one  breaks,  the  other  will  carry  the  load.) 

187.  A  belt  is  carried  by  a  36  in.  pulley  running  at  150  R.  P.  M.     The 
effective  pull  in  the  belt  is  240  Ib.     Find  the  horse-power. 

188.  A  pumping  engine  lifts  92,500  gallons  of  water  every  hour  to  a 
height  of  150  ft.     What  is  the  horse-power  of  the  engine? 

189.  If  a  freight  elevator  and  its  load  weigh  5000  Ib.,  what  horse-power 
must  be  exerted  to  raise  the  elevator  at  a  rate  of  2  ft.  per  second? 

190.  A  touring  car  is  travelling  on  a  level  road  at  a  rate  of  45  miles  an 
hour.     If  it  is  shown  by  actual  test  that  a  force  of  200  Ib.  is  required  to 
maintain  this  rate  of  speed,  what  horse-power  must  the  engine  deliver  at 
the  wheels? 


CHAPTER  XVII 
HORSE-POWER  OF  ENGINES 

109.  Steam  Engines. — In  the  last  chapter,  the  meaning  of  the 
term  horse-power  was  explained  and  its  application  to  belting 
was  discussed.  We  will  now  take  up  the  calculations  of  the 
horse-powers  of  steam  and  gas  engines. 

One  horse-power  was  given  as  the  ability  to  do  33,000  foot- 
pounds of  work  in  1  minute.  From  this  we  see  that  the  best  way 
to  get  the  horse-power  of  any  engine  is  to  find  out  how  many 
foot-pounds  of  work  it  does  in  1  minute  and  then  to  divide  the 
number  of  foot-pounds  delivered  in  a  minute  by  33,000. 

Let  us  study  the  action  of  the  steam  in  the  cylinder  of  the 
ordinary  double-acting  steam  engine.  In  Fig.  70  is  shown  a 


FIQ.  70. 

section  of  a  very  simple  boiler  and  engine.  We  find  that  steam 
enters  one  end  of  the  cylinder  behind  the  piston  and  pushes  the 
piston  toward  the  other  end  of  the  cylinder.  Meanwhile,  the 
valve  is  moved  to  the  other  end  of  the  valve  chest.  The  opera- 
tion is  then  reversed  and  the  piston  is  pushed  back  to  the  starting- 
point.  It  has  thus  made  two  strokes,  or  one  revolution.  The 
steam  pressure  on  the  piston  is  not  the  same  at  all  points  in  the 
stroke,  but  varies  according  to  the  action  of  the  valve  in  cutting 
off  the  admission  of  steam  into  the  cylinder.  However,  it 

145 


146  SHOP  ARITHMETIC 

is  possible  to  obtain  the  average  or  "mean  effective  pres- 
sure" per  square  inch  during  a  stroke,  and,  if  we  multiply 
this  by  the  piston  area  in  square  inches,  we  will  have  the  average 
total  pressure  or  force  exerted  during  one  stroke.  Now  reduce 
the  length  of  the  stroke  to  feet  and  multiply  this  by  the  total 
pressure  just  found,  and  we  have  the  number  of  foot-pounds  of 
work  done  during  one  stroke.  This  result,  when  multiplied  by 
the  number  of  working  strokes  per  minute,  gives  the  foot-pounds 
per  minute  and  this  divided  by  33,000  gives  the  horse-power. 
The  following  are  the  symbols  generally  used  : 

H.  P.  —  Horse-power. 

P  =  Mean  pressure  in  pounds  per  square  inch. 
A  =  Area  of  piston  in  square  inches. 
L  =  Length  of  stroke  in  feet  . 
N  =  Number  of  working  strokes  per  minute. 
PX  A  =  Total  pressure  on  piston. 
PX  A  xL  =  it.  Ib.  of  work  done  per  stroke. 
PXA  XLX  N  —  it.  Ib.  of  work  done  per  minute,  and  hence 

PXAXLXN 
=        33000      "  or'  as  usually  wntten> 

PXLXAXN 

-  .     In  the  latter  form,  the  letters  in  the  numerator 
33000 

spell  the  word  Plan  and  the  formula  is  thus  easily  remembered. 
In  the  common  steam  engine,  there  are  two  working  strokes 
for  every  revolution  of  the  engine,  that  is,  the  engine  is  \vhat  is 
called  double  acting,  and  N  is  twice  the  revolutions  per  minute. 
A  few  steam  engines,  like  the  vertical  Westinghouse  engine,  are 
single  acting  and,  hence,  have  only  one  working  stroke  of  each 
piston  per  revolution.  Unless  otherwise  stated,  it  will  be  as- 
sumed in  working  problems  that  a  steam  engine  is  double  acting. 

Example  : 

Find  the  horse-power  of  a  32  in.  by  54-in.  steam  engine  running  at 
94  R.  P.  M.  with  an  M.  E.  P.  (Mean  Effective  Pressure)  of  60  Ib. 

Note.  —  In  giving  the  dimensions  of  an  engine  cylinder,  the  first  number 
represents  the  diameter  and  the  second  number  the  stroke. 
P  =  60  Ib. 

L  =  54  in.  =4^  ft. 

A  =Area  of  32  in.  piston  =  804.  25  sq.  in. 
N  =  Number  of  strokes  per  minute  =  94  X  2  =  188 
H   p  _-PxLxAxJV 


33000 
60X4.5X804.25X188 


1237+  Jiorse-power,  Answer. 


HORSE-POWER  OF  ENGINES  147 

Notice  particularly  that  the  area  of  the  piston  is  expressed  in 
square  inches,  because  the  pressure  is  given  in  pounds  per  square 
inch;  but  that  the  stroke  is  reduced  to  feet  because  we  measure 
work  in  foot-pounds  and,  consequently,  must  express  in  feet  the 
distance  which  the  piston  moves. 

If  an  engine  has  more  than  one  cylinder,  the  horse-power  of 
each  can  be  calculated  and  the  results  added;  or,  if  the  cylinders 
are  arranged  to  do  equal  amounts  of  work,  we  can  find  the  horse- 
power of  one  cylinder  and  multiply  this  by  the  number  of 
cylinders. 

The  mean  effective  pressure  can  be  obtained  for  any  engine 
by  the  use  of  a  device  called  an  "indicator,"  which  draws  a 
diagram  showing  just  what  the  pressure  is  in  the  cylinder  at  each 
point  in  the  stroke.  From  this  diagram,  we  can  calculate  the 
average  or  mean  effective  pressure  for  the  stroke.  This  pressure 
must  not  be  confused  with  the  boiler  pressure  or  the  pressure  in 
the  steam  pipe.  For  instance,  when  the  steam  comes  from  the 
boiler  to  the  engine  at  100  Ib.  pressure,  the  mean  pressure  in  the 
cylinder  will  not  be  100  Ib.,  as  it  would  be  very  wasteful  to  use 
steam  from  the  boiler  for  the  full  stroke.  Instead,  the  M.  E.  P. 
(Mean  Effective  Pressure)  will  be  from  20%  to  85%  of  the  boiler 
pressure  depending  on  the  type  of  the  engine  and  the  load  it  is 
carrying.  Horse-power  calculated  as  explained  here  is  called 
Indicated  Horse-power  because  an  indicator  is  used  to  determine 
it.  The  indicated  horse-power  represents  the  power  delivered 
to  the  piston  by  the  steam. 

110.  Gas  Engines. — The  most  common  type  of  gas  or  gasoline 
engine  works  on  what  is  called  the  four  stroke  cycle.  Such 
an  engine  is  called  a  four-cycle  engine.  Fig.  71  shows  in  four 
views  the  operation  of  such  an  engine.  Four  strokes,  or  two 
revolutions,  are  required  for  each  explosion  that  occurs  in  the 
cylinder.  Consequently,  in  calculating  the  horse-power  of  a 
single  cylinder  gas  engine,  the  number  of  working  strokes  (or  N 
in  the  horse-power  formula)  is  one-half  of  the  R.  P.  M.  There 
is  another  type  of  gasoline  engine  called  the  two-cycle  engine. 
A  single  cylinder  two-cycle  engine  has  one  working  stroke  for 
each  revolution  of  the  crank  shaft  and  N  is  therefore  the  same 
as  the  number  of  R.  P.  M. 

The  mean  effective  pressure  of  a  gas  engine  is  from  40  to  100 
Ib.  per  square  inch,  depending  chiefly  on  the  fuel  used.  For 


148 


SHOP  ARITHMETIC 


gasoline  or  natural  gas  or  illuminating  gas  it  is  usually  between 
80  and  90  Ib.  per  square  inch. 


2.  Compression 


4-.  Exttausr 


Fio.  71 


Example  : 

What  horse-power  could  be  delivered  by  a  single  cylinder  5  in. 
by  8  in.  four-cycle  gasoline  engine  running  450  R.  P.  M.? 
Note.  —  Use  a  value  of  P  =  80  Ib.  per  square  inch. 
P  =  80 


A  =  .  7854  X52  =  19.6  sq.  in. 


Then,  H.  P.  •• 


33000 
80x|xl9.6x225 

O 

" 33000 


-  =  7.13  horse-power,  Answer. 


HORSE-POWER  OF  ENGINES  149 

111.  Air  Compressors.  —  An  air  compressor  is  like  a  double 
acting  steam  engine  in  appearance;  but,  instead  of  delivering  up 
power,  it  requires  power  from  some  other  source  to  run  it.  This 
power  is  stored  in  the  air  and  later  is  recovered  when  the  air  is 
used.  An  air  compressor  takes  air  into  the  cylinder,  raises  its 
pressure  by  compressing  it,  and  then  forces  it  into  the  air  line  or 
the  storage  tank.  In  calculating  the  horse-power  of  a  com- 
pressor, the  same  formula  can  be  used  as  for  a  steam  engine. 
The  value  of  P  to  use  is  not  the  pressure  to  which  the  air  is  raised, 
but  is  the  average  or  mean  pressure  during  the  stroke.  It  is 
usually  somewhat  less  than  half  the  final  air  pressure;  for  ex- 
ample, when  an  air  compressor  is  delivering  air  at  80  Ib.  pressure, 
the  mean  pressure  on  the  piston  is  about  33  Ib. 

Most  air  compressors  are  double  acting,  though  there  are  many 
small  single  acting  ones. 

Example  : 

A  double  acting  12  in.  by  14  in.  air  compressor  is  running  150 
R.  P.  M.  It  is  supplying  air  at  100  Ib.  and  the  mean  pressure  in  the  cylinder 
is  37  Ib.  per  square  inch.  Calculate  the  horse-power  necessary  to  run  it. 

P  =  37  Ib. 


A  =  Area  of  12  in.  piston  =  113.1  sq.  in. 
N  =  Strokes  =  150  X  2  =  300  per  minute. 

ThPn    ff   P 
Then,  H.P.  - 


6       110 

In  this  case  12  appears  in  the  denominator  in  order  to  reduce  the  14  inches 
to  feet. 

112.  Brake  Horse-power.  —  The  Brake  Horse-power  of  an 
engine  is  the  power  actually  available  for  outside  use.  It, 
therefore,  is  equal  to  the  indicated  horse-power  minus  the  power 
lost  in  friction  in  the  engine.  Brake  horse-power  can  be  readily 
determined  by  putting  a  brake  on  the  rim  of  the  fly-wheel  and 
thus  absorbing  and  measuring  the  power  actually  delivered. 
Fig.  72  shows  such  a  brake  arranged  for  use.  This  form  is 
known  as  the  "Prony  Brake."  It  consists  of  a  steel  or  leather 
band  carrying  a  number  of  wooden  blocks.  By  tightening  the 
bolt  at  A,  the  friction  between  the  blocks  and  the  rim  of  the  wheel 
can  be  varied  at  will.  The  corresponding  pull  which  this  friction 
gives  at  a  distance  R  ft.  from  the  shaft  is  weighed  by  a 


150 


SHOP  ARITHMETIC 


platform  scale  or  spring  balance.  From  the  scale  reading  must 
be  deducted  the  weight  due  to  the  unbalanced  weight  of  the 
brake  arms,  which  can  be  determined  by  reading  the  scales  when 
the  brake  is  loose  and  the  engine  is  not  running.  If  an  engine 
is  capable  of  maintaining  a  certain  net  pressure  W  on  the  scale, 
and  meanwhile  maintains  a  speed  of  N  revolutions  per  minute, 
we  can  readily  see  that  this  is  equivalent  to  an  effective  belt  pull 
of  W  pounds  on  a  pulley  of  radius  R  running  at  N  revolutions; 
or  it  can  be  considered  as  being  equivalent  to  raising  a  weight 


FIG.  72. 

equal  to  W  by  means  of  a  rope  wound  around  a  pulley  of  radius 
R  turning  at  N  revolutions  per  minute.  This  weight  would  be 
lifted  at  the  rate  of 

3.1416 X2XRXN  ft.  per  minute 
and  the  brake  horse-power  will  be 

'      <P'=  33,000 

The  brake  and  wheel  rim  will  naturally  get  hot  during  a  test,  as 
all  of  the  work  done  by  the  engine  is  transformed  back  into  heat 
at  the  rubbing  surfaces  of  the  pulley  rim  and  the  brake.  It  is 
necessary  to  keep  a  stream  of  water  playing  on  the  rim  to  remove 
this  heat  and  it  is  best  to  have  special  brake  wheels  for  testing. 
These  have  thin  rims  and  inwardly  extending  flanges  on  the 
rims  so  that  a  film  of  water  can  be  maintained  on  the  inner  sur- 
face of  the  rim. 


HORSE-POWER  OF  ENGINES  151 

Example  : 

Suppose  that,  at  the  time  of  testing  the  5x8  gas  engine  in  article 
110,  we  also  determined  the  brake  horse-power  by  means  of  a  Prony  brake 
having  a  radius  of  3  ft.  and  that  a  net  pressure  of  22  Ib.  was  exerted  on  the 
scales  (the  speed  of  the  engine  was  450  II.  P.  M.).  Let  us  calculate  the  brake 
horse-power. 

Explanation:    Our  data  is  equi- 
•}  l41fiv9VW4{in-84R2  valent  to  that  of  hoisting  a  weight 

22X8482-186  604?t   Ib  °f  22  lb"  ^  a  r°Pe  windin^  uPon 

K  «k    J    c  a  pulley  of  3  ft.  radius  turning  at 

186,604  +  33,000  =  5.65,  Answer.         ^   ^p    M      The  22  Jb    wefght 

would    rise  8482   ft.    per  minute 

and    the   work   done   per  minute  would  be  22  Ib.X  8482  ft.  =  186604  foot- 
pounds per  minute.     Hence,  the  brake-horse  power  of  the  engine  is  5.65. 

113.  Frictional  Horse-power.  —  If  this  engine  gave  7.13  indi- 
cated H.   P.   (I.   H.  P.),  but  the  power  available  at  the  fly- 
wheel was  only  5.65,  it  stands  to  reason  that  the  difference,  or 
1.48    H.    P.,    was   lost   between   the    cylinder    and    fly-wheel. 
The  explanation  is  that  this  power  is  expended  in  simply  over- 
coming the  friction  of  the  engine;  and  this  horse-power  is,  there- 
fore, called  the  Frictional  Horse-power.     At  zero  brake  horse- 
power, the  entire  I.  H.  P.  is  used  in  overcoming  friction. 

114.  Mechanical  Efficiency.  —  The  ratio  of  the  Brake  Horse- 
power   to    the    Indicated    Horse-power    gives    the    mechanical 
efficiency,  meaning  the  efficiency  of  the  mechanism  in  trans- 
mitting the  power  through  it  from  piston  to  fly-wheel.     This  is 
usually  expressed  in  per  cent.     In  the  case  of  the  engine  of 
which  we  figured  the  I.  H.  P.  and  B.  H.  P.,  the  mechanical 
efficiency  was 

K    «^ 

£=          =  .792  =  79.2% 


The  mechanical  efficiency  of  a  gas  engine  is  lower  than  that  of  a 
steam  engine  on  account  of  the  idle  strokes  which  use  up  work 
in  friction  while  no  power  is  being  generated,  but  at  full  load  a 
well  built  gas  engine  should  show  over  80  per  cent,  mechanical 
efficiency.  The  mechanical  efficiency  of  a  steam  engine  should 
be  above  90%  at  full  load. 

PROBLEMS 

191.  The  cage  in  a  mine  weighs  2200  Ib.  and  the  load  hoisted  is  3  tons, 
The  hoisting  speed  is  20  ft.  per  second.     Calculate  horse-power  necessary. 
allowing  25%  additional  for  friction  and  rope  losses. 

192.  A  10  in.  by  12  in.  air  compressor  runs  150  R.  P.  M.     The  M.  E.  P. 
is  30  Ib.     Calculate  the  horse-power  required  to  run  it. 

193.  A  pump  lifts  2000  gallons  of  water  per  minute  into  a  tank  150  ft. 
above  it.     Find  the  horse-power  of  the  pump. 


152  SHOP  ARITHMETIC 

194.  Find  the  horse-power  of  a  10  in.  by  12  in.  steam  engine  running 
250  R.  P.  M.  with  a  M.  E.  P.  of  60  Ib. 

195.  What  will  be  the  horse-power  of  a  single  cylinder,  four  cycle  gas 
engine  with  the  following  data: 

Size  of  cylinder,  12  in.  by  16  in. 

Rev.  per  minute,  225 

Mean  effective  pressure,  78  Ib.  per  square  inch? 

Numhrer  of  working  strokes  =  J  of  the  number  of  revolutions. 

196.  A  body  can  do  as  much  work  in  descending  as  is  required  to  raise  it. 
Knowing  this  fact,  calculate  the  horse-power  that  could  be  developed  by  a 
water-power  which  discharges  800  cu.  ft.  of  water  per  second  from  a  height 
of  13.6  ft.,  assuming  that  25%  of  the  theoretical  power  is  lost  in  the  wheel 
and  in  friction. 

197.  What  would  be  the  brake  horse-power  of  a  steam  engine  which 
exerted  a  net  pressure  of  100  Ib.  on  the  scales,  at  a  radius  of  4  ft.,  when 
running  at  250  R.  P.  M.? 

198.  How  many  foot-pounds  of  work  per  hour  would  be  obtained  from 
a  60  H.  P.  engine? 

199.  A  centrifugal  pump  is  designed  to  pump  3000  gallons  of  water  per 
minute  to  a  height  of  70  ft.     If  the  efficiency  of  the  pump  is  60%,  what 
horse-power  will  be  required  to  drive  it? 

200.  The  pump  of  problem  199  is  to  run  1500  R.  P.  M.  and  is  to  be  belt- 
driven  from  a  48  in.  pulley  on  a  high  speed  automatic  engine,  running  275 
R.  P.  M.     What  should  be  the  diameter  and  width  of  fane  of  the  pulley  on 
the  pump,  if  the  pulley  is  to  be  1  in.  wider  than  the  belt? 


CHAPTER  XVIII 
MECHANICS  OF  FLUIDS 

116.  Fluids. — Nearly  every  shop  of  any  size  contains  some 
devices  which  are  operated  by  water  or  air  pressure,  so  every  up- 
to-date  mechanic  should  have  a  knowledge  of  how  these  machines 
work. 

A  Fluid  is  any  substance  which  has  no  particular  form,  but 
always  shapes  itself  to  the  vessel  which  contains  it.  Water,  oil, 
air,  steam,  gas — all  are  fluids.  In  some  ways  water,  oil,  and 
similar  substances  are  different  from  the  lighter  substances — air, 
steam,  etc.  To  separate  these,  we  give  the  name  of  Liquids  to 
such  substances  as  water  and  oil;  while  air,  steam,  etc.,  are  given 
the  general  name  of  Gases.  In  some  respects  liquids  and  gases 
are  alike  and  in  others  they  are  different.  The  chief  difference  is 
that  liquids  have  definite  volumes;  they  cannot  be  compressed 
or  expanded  any  visible  amount,  while  gases  can  be  readily 
compressed  or  expanded  to  almost  any  extent.  For  all  practical 
purposes  we  can  say  that  liquids  cannot  be  compressed.  The 
third  form  of  matter — Solids — needs  no  explanation.  The 
differences  in  these  three  forms  can  be  stated  as  follows: 

A  Solid  has  a  definite  shape  and  volume. 

A  Liquid  has  no  definite  shape,  but  has  a  definite  volume. 

A  Gas  has  neither  a  definite  shape  nor  volume. 

There  are  some  substances  that  exist  in  states  in  between  the 
solid  and  the  liquid  form.  Among  these  are  tar,  glue,  putty, 
gelatine,  etc. 

116.  Specific  Gravity. — By  Specific  Gravity  of  a  substance  we 
mean  its  relative  weight  as  compared  with  the  same  volume  of 
water.  Thus  we  say  that  the  specific  gravity  of  cast  iron  is  7.21, 
meaning  that  cast  iron  is  7.21  times  as  heavy  as  water.  A  cubic 
foot  of  water  weighs  62.4  Ib.  and  a  cubic  foot  of  cast  iron  weighs 
about  450  Ib.  The  quotient  *2™  =7.21  is  the  specific  gravity 
of  the  iron. 

In  many  hand  books  we  find  tables  of  specific  gravities  and, 
when  we  wish  to  get  the  actual  weight  per  cubic  inch  or  per  cubic 
H  153 


154. 


SHOP  ARITHMETIC 


foot  of  some  substance,  we  must  multiply  the  weight  of  the  unit 
of  water  by  the  specific  gravity  of  the  substance. 

Example : 

The  specific  gravity  of  alcohol  is  .8.     How  many  pounds  would 
there  be  to  a  gallon  of  alcohol? 

One  gallon  of  water  =8J  Ib. 

One  gallon  of  alcohol  =  .8  X8J  =  6f  Ib.,  Answer. 

If  a  substance  has  a  specific  gravity  less  than  1,  it  will  float  in 
water,  because  it  is  lighter  than  the  same  volume  of  water.  If 
the  specific  gravity  is  greater  than  1,  the  substance  is  heavier 
than  water  and  will  sink.  A  substance  that  will  float  in  water 
may  sink  in  some  other  liquid  if  it  has  a  greater  specific  gravity 
than  the  liquid  in  question.  For  example,  a  piece  of  apple-wood 
will  float  in  water  but  will  sink  when  placed  in  gasoline,  the 
specific  gravity  of  the  wood  being  about  .76  and  that  of  gasoline 
about  .71.  A  piece  of  iron  will  sink  in  water  but  will  float  when 
placed  in  mercury  (quick  silver) ,  the  specific  gravity  of  mercury 
being  13.6  and  that  of  iron  about  7.21. 

117.  Transmission  of  Pressure  Through  Fluids. — One  of  the 
most  useful  properties  of  all  fluids  is  the  ability  to  transmit 


FIG.  73. 

pressure  in  all  directions.  If  we  have  a  vessel  filled  with  water, 
as  shown  in  Fig.  73,  and  apply  a  pressure  to  the  water  by  means 
of  a  piston,  as  shown,  this  pressure  will  be  transmitted  through 
the  water  in  all  directions.  If  the  sides  of  the  vessel  are  flat, 
they  will  bulge  out,  showing  that  there  is  a  pressure  on  the  sides; 
and  if  the  piston  is  loose,  the  water  will  escape  upward  around 
it,  showing  that  there  is  a  pressure  in.this  direction  also. 


MECHANICS  OF  FLUIDS 


155 


If  the  total  force  on  the  piston  is  W  Ib.  and  the  area  on 
the  bottom  of  the  piston  is  A  sq.  in.,  then  there  will  be  a  pressure 

W 
of  -j-  Ib.  exerted  on  each  square  inch  of  the  water  beneath  the 

piston.     This  pressure  will  be  transmitted  equally  in  all  direc- 
tions and  the  pressure  on  each  square  inch  of  the  top,  sides,  and 

W 

bottom  of  the  vessel  will  be  -j-  Ib. 

Example  : 

If  the  piston  of  Fig.  73  is  6  in.  in  diameter,  and  has  a  total  weight 
of  1000  Ib.,  what  would  be  the  water  pressure  per  square  inch? 


W 


1000 
.7854  ><6* 


p 

Explanation:  As  the  piston 
rests  on  the  water,  the  pressure 
°f  the  water  on  the  bottom  of 

1000  the  piston  must  be  sufficient  to 

-5^-==—  35.4  Ib.  per  sq.  in.,  Answer,    support  the  weight.     The  area 

of  the  bottom  is  28.27  sq.  in., 

1000 
and  the  pressure  on  each  sq.  in.   will    be  ^nvr  or  35.4  Ib.  persquare  inch. 

Zo.Z  I 

This  pressure  is  transmitted  throughout  the  water  and  is  exerted  by  it  with 
equal  force  in  all  directions. 

118.  The  Hydraulic  Jack.  —  This  property  of  water  of  trans- 
mitting pressure  in  any  direction  is  made  use  of  in  many  ways. 
The  same  property  is,  of  course,  common  to  other  fluids  such  as 


FIG.  74. 

oil,  air,  etc.  Wherever  we  find  a  powerful,  slow-moving  force 
required  in  a  shop,  we  usually  find  some  hydraulic  machine. 
(The  word  "  hydraulic  "  refers  to  the  use  of  water  but  it  is  often 
applied  to  machines  using  any  liquid — water,  oil,  or  alcohol.) 
Fig.  74  shows  the  principle  of  all  these  hydraulic  machines. 


156  SHOP  ARITHMETIC 

A  small  force  P  is  exerted  on  a  small  piston  and  this  produces  a 
certain  pressure  in  the  water.  This  pressure  is  transmitted  to 
the  larger  cylinder  where  the  same  pressure  per  square  inch  is 
exerted  on  the  under  side  of  the  large  piston.  If  the  large  piston 
has  100  times  the  area  of  the  small  piston,  the  weight  supported 
(W)  will  be  100  times  P.  If  the  water  pressure  produced  by  P 
is  100  Ib.  per  square  inch  and  the  large  piston  has  an  area  of  100 
sq.  in.,  then  the  weight  W  that  can  be  raised  will  be  10,000  Ib. 

Like  the  lever,  the  jackscrew,  and  the  pulley,  this  increase  in 
force  is  obtained  only  by  a  decrease  in  the  distance  the  weight  is 
moved.  The  work  done  on  the  small  piston  is  theoretically  the 
same  as  the  work  obtained  from  the  large  piston.  For  example, 
suppose  that  the  large  piston  has  100  times  the  area  of  the  small 
one  and  we  shove  the  small  piston  down  1  in. ;  the  water  that  is 
thus  pushed  out  of  the  small  cylinder  will  have  to  spread  out 
over  the  entire  area  of  the  large  piston;  the  large  piston  will, 
therefore,  be  raised  only  one  one-hundredth  of  the  distance  that 
the  small  piston  was  moved.  Thus,  we  have,  here  also,  an  applica- 
tion of  the  law  that  the  work  put  into  a  machine  is  equal, 
neglecting  friction,  to  the  work  done  by  it. 

Force  X  distance  moved  =  weight  X  distance  raised. 

The  Mechanical  Advantage  of  such  a  machine  will  be  seen  to 
be  the  ratio  of  the  areas  of  the  pistons.  In  the  case  just  men- 
tioned, the  ratio  of  the  areas  of  the  pistons  was  100:1;  hence, 
the  mechanical  advantage  was  100. 

In  Fig.  74,  the  motion  that  can  be  given  to  W  is  very  limited, 
but  by  using  a  pump  with  valves,  instead  of  the  simple  plunger 
P,  we  can  continue  to  force  water  into  the  large  cylinder  and 
thus  secure  a  considerable  motion  to  W. 

Fig.  75  shows  a  common  form  of  hydraulic  jack  which  operates 
on  this  principle.  The  top  part  contains  a  reservoir  for  the 
liquid,  and  also  has  a  small  pump  operated  by  a  hand  lever  on 
the  outside  of  the  jack.  By  working  the  lever,  the  liquid  is 
pumped  into  the  lower  part  of  the  jack  between  the  plunger  and 
the  casing,  thus  raising  the  load.  The  load  may  be  lowered  by 
slacking  the  lowering  screw  Y.  This  opens  a  passage  to  the 
reservoir,  and  the  load  on  the  jack  forces  the  liquid  to  flow  back 
through  this  passage  to  the  reservoir. 

In  calculating  the  mechanical  advantage  of  a  hydraulic  jack, 
we  must  consider  the  mechanical  advantage  of  the  lever  which 


MECHANICS  OF  FLUIDS 


157 


operates  the  pump  as  well  as  the  advantage  due  to  the  relative 
sizes  of  the  pump  and  the  ram. 


Fia.  75. 


Example : 

If  the  ram  of  Fig.  75  is  3  in.  in  diameter  and  the  pump  is  1  in.  in 
diameter,  while  the  lever  is  15  in.  long  and  is  connected  to  the  pump  at  a 
distance  of  1 J  in.  from  the  fulcrum,  what  is  the  mechanical  advantage  of  the 
entire  jack? 

Explanation:  The  areas  of  the  ram  and  pumps 
are  as  9:1,  hence  their  mechanical  advantage  is  9. 
The  lever  has  a  mechanical  advantage  of  10. 
Hence,  that  of  the  whole  jack  is  9X10  =  90,  and 
a  force  applied  at  the  end  of  the  lever  would  be 
multiplied  90  times.  This  force  would,  however, 
move  through  a  distance  90  times  as  great  as  the 
distance  the  load  would  be  raised. 


•7854X32_9 
.7854  Xl1""!" 

9X10  =  90,  Answer. 


158 


SHOP  ARITHMETIC 


The  hydraulic  jack  has  usually  an  efficiency  of  over  70%  and 
is,  therefore,  a  much  more  efficient  lifting  device  than  the  jack 
screw.  A  mixture  containing  one-third  alcohol  and  two-thirds 
water  should  be  used  in  jacks.  The  alcohol  is  added  to  prevent 
freezing. 

119.  Hydraulic  Machinery. — In  the  shop,  we  often  find  water 
pressure  used  to  operate  presses,  punches,  shears,  riveters, 
hoists,  and  sometimes  elevators.  These  machines  are  seldom 
operated  by  hand  power  but  have  water  supplied  under  pressure 
from  a  central  pumping  plant.  The  admission  of  the  water  and 
the  consequent  motion  of  the  machine  is  controlled  by  hand 
operated  valves.  Most  of  these  hydraulic  machines  are  used 
where  tremendous  forces  are  required.  Therefore,  very  high 
water  pressures  are  used,  occasionally  as  high  as  3000  Ib.  per 
square  inch.  1500  Ib.  per  square  inch  is  a  common  working 
pressure  for  hydraulic  machines. 


FIG.  76. 


FIG.  77. 


Fig.  76  shows  a  press  operated  by  hydraulic  pressure.  It  will 
be  noticed  that  the  movable  head  is  connected  to  two  pistons — 
a  large  one  for  doing  the  work  on  the  down  stroke,  and  a  smaller 
one  above,  used  only  for  the  idle  or  return  stroke  of  the  press. 

120.  Hydraulic  Heads. — Quite  often  we  use  a  high  tower  or 


MECHANICS  OF  FLUIDS  159 

tank  to  secure  a  water  pressure,  or  we  make  use  of  some  natural 
source  of  water  which  is  at  some  elevation.  This  is  most  often 
seen  in  the  water  supplies  for  towns  and  cities.  Water  tanks 
are  put  upon  the  roofs  of  some  factories  for  the  same  purpose. 
The  higher  the  tank,  the  greater  will  be  the  pressure  which  it  will 
maintain  in  the  system.  Let  Fig.  77  represent  such  a  system. 
The  water  at  the  bottom  has  the  weight  of  a  column  of  water 
h  ft.  high  to  support  and,  consequently,  will  be  under  a  pressure 
equal  to  the  weight  of  this  column  of  water.  A  volume  of  water 
1  in.  square  and  1  ft.  high  weighs  .434  lb.,  so  the  pressure  per 
square  inch  at  the  base  of  the  column  in  Fig.  77  will  be  .434  X  h. 
Notice  particularly  that  the  shape  and  size  of  the  tank  has  no 
influence  on  the  pressure,  it  being  used  merely  for  storage  and  to 
keep  the  pressure  from  falling  too  fast  if  the  water  is  drawn  off. 
The  water  in  the  tank  on  either  side  of  the  outlet  is  supported  by 
the  bottom  of  the  tank  and  has  no  effect  on  the  pressure  in  the 
pipe.  The  pressure  at  the  bottom  of  the  pipe  would  be  the  same 
if  the  pipe  alone  extended  up  to  the  height  h  without  the  tank. 
Also  the  size  of  the  pipe  has  no  effect  on  the  pressure  per  square 
inch.  The  water  in  a  large  pipe  will  weigh  more  than  in  a  small 
pipe,  but  the  pressure  will  be  spread  over  a  larger  area  and  if, 
the  heights  are  the  same,  the  pressure  per  square  inch  will  be  the 
same. 

In  pumping  water  to  an  elevated  tank  or  reservoir,  the  pressure 
required  per  square  inch  is  also  determined  in  the  same  manner 
and  is  .434  times  the  height  to  which  the  water  is  raised,  plus  an 
allowance  for  friction  in  the  pipes.  Thus,  to  pump  water  to  a 
height  of  100  ft.  requires  a  pressure  somewhat  greater  than  .434  X 
100  =  43.4  lb.  per  square  inch. 

121.  Steam  and  Air. — Steam  and  air  are  likewise  used  to  pro- 
duce pressures  in  shop  machinery.  Being  more  elastic  than 
water,  they  are  preferred  where  the  machines  are  to  be  operated 
quickly.  Devices  using  air  are  called  "pneumatic  appliances," 
among  the  most  common  of  which  are  pneumatic  drills,  hammers, 
and  hoists.  The  air  for  operating  these  is  supplied  by  air  com- 
pressors located  in  the  power  house.  These  take  the  air  from  out 
of  doors  and  compress  it  into  a  smaller  volume.  The  resistance 
of  the  air  to  this  compression  causes  it,  in  its  effort  to  escape,  to 
exert  a  pressure  on  the  walls  of  the  tank  or  pipe  containing  it. 
The  more  the  air  is  compressed,  the  greater  is  the  pressure  exerted 
by  it.  The  air  pressure  used  in  shop  work  is  usually  about 


160 


SHOP  ARITHMETIC 


80  Ib.  per  square  inch.  The  air  is  conducted  through  pipes  and 
hose  to  the  point  where  it  is  to  be  used  and  there  allowed  to 
exert  its  pressure  on  the  piston  of  the  appliance  which  is  to  be 
driven. 


PROBLEMS 

201.  The  specific  gravity  of  Lignum  Vitse  (a  hard  wood)  is  1.328. 
this  wood  float  or  will  it  sink  in  water? 


Will 


® 


TTT 


£ 


fit 


FIG.  78. 


FIG.  79. 


202.  What  weight  on  the  small  piston  of  Fig.  74  would  support  a  weight 
of  30,000  Ib.  on  the  large  piston  if  the  small  piston  is  1  in.  in  diameter  and 
the  large  one  12  in.  in  diameter? 

203.  If  a  hydraulic  press  works  with  a  water  pressure  of  1500  Ib.  per 
square  inch,  what  must  be  the  diameter  of  the  ram  if  a  total  pressure  of 
75,000  Ib.  is  to  be  produced? 

204.  If  the  air  hoist  of  Fig.  78  has  a  cylinder  10  in.  in  diameter  inside, 
and  the  piston  rod  is  1J  in.  in  diameter,  and  an  air  pressure  of  80  Ib.  per 
square  inch  is  exerted  on  the  bottom  of  the  piston,  what  weight  can  be 
lifted  by  the  hoist? 

205.  If  a  city  wishes  to  maintain  a  water  pressure  of  80  Ib.  per  square 
inch  at  their  hydrants,  how  high  above  the  streets  must  be  the  water  level 
in  the  stand  pipe? 


MECHANICS  OF  FLUIDS 


161 


206.  Fig.  79  shows  the  principle  of  one  form  of  hydraulic  elevator,  the 
car  being  fastened  directly  to  a  long  ram  which  is  raised  by  water  pressure. 
The  weight  of  the  ram  and  car  are  partially  balanced  by  a  counterweight. 
If  this  elevator  is  operated  with  water  from  the  city  mains  at  80  Ib.  pressure 
per  square  inch  and  the  ram  has  a  diameter  of  10  in.,  what  load  can  be 
lifted  allowing  30%  for  losses  in  friction,  etc.? 


FIQ.  80. 


Fia.  SI. 

207.  If  a  steam  pump,  such  as  shown  in  Fig.  80,  has  a  12  in.  steam  piston 
and  an  8  in.  water  piston,  what  water  pressure  can  be  produced  with  a  steam 
pressure  of  90  Ib.  per  square  inch?     How  many  gallons  would  be  pumped 
per  minute  when  the  pump  is  running  at  100  strokes  per  minute,  the  stroke 
of  the  pump  being  12  in.?     (1  gallon  =  231  cu.  in.) 

208.  What  water  pressure  must  a  pump  be  capable  of  producing  in  order 
to  force  the  water  to  a  reservoir  at  an  elevation  of  300  ft.  above  the  pump? 

15 


162  SHOP  ARITHMETIC 

209.  A  gravity  oiling  system  has  an  oil  tank  placed  10  ft.  above  the 
bearings  to  be  lubricated.     The  tank  is  connected  by  small  tubes  to  the 
various  bearings.    If  the  specific  gravity  of  the  oil  is  .88,  what  pressure  will 
the  oil  have  at  the  bearings? 

210.  In  Fig.  81  we  have  a  hoist  operated  by  a  hydraulic  ram  in  the  top 
of  the  crane  post.     The  motion  of  the  ram  is  multiplied  by  the  system  of 
pulleys  shown  in  the  figure.     What  size  must  the  ram  be  that  a  load  of 
10,000  Ib.  can  be  lifted  with  a  water  pressure  of  72  Ib.  per  square  inch,  the 
efficiency  of  the  whole  apparatus  being  70%? 


CHAPTER  XIX 
HEAT 

122.  Nature  of  Heat. — Some  of  the  effects  of  heat  are  very- 
useful  in  shop  work  and  every  mechanic  should  know  something 
of  the  nature  of  heat  and  of  the  laws  which  govern  its  applica- 
tions to  shop  work. 

Heat  is  a  form  of  energy;  that  is,  it  is  capable  of  doing  work. 
This  we  see  amply  illustrated  in  the  steam  engine  and  the  gas 
engine,  where  heat  is  used  in  producing  work.  The  steam  engine 
uses  heat  which  has  been  imparted  to  the  steam  in  the  boiler. 
Part  of  the  heat  of  the  steam  is  changed  to  work  in  the  engine  and 
the  rest  is  rejected  in  the  exhaust.  Heat  is  not  a  substance  as 
was  formerly  supposed — it  cannot  be  weighed  and  cannot  exist 
by  itself. 

It  is  always  found  in  some  substances.  We  generally  get  heat 
by  burning  some  fuel  such  as  coal,  wood,  gas,  or  oil.  In  burning, 
the  fuel  unites  with  oxygen,  one  of  the  constituents  of  air,  and 
this  process,  called  combustion,  generates  the  heat.  We  cannot 
get  heat  by  this  process,  therefore,  without  air.  No  fuel  will 
burn  without  a  supply  of  air,  and  as  soon  as  we  shut  off  the  air 
from  a  fire,  combustion  stops  and  no  more  heat  is  generated. 
A  fire  may  continue  to  give  off  heat  for  some  time  after  the  air  is 
cut  off,  but  this  heat  comes  from  the  cooling  of  the  hot  fuel  in 
the  fire.  Of  the  heat  generated  during  combustion,  some  of  it 
goes  through  the  furnace  walls  to  the  surrounding  air;  some  goes 
to  heat  up  the  bed  of  coals  and  any  object  that  may  be  placed  in 
the  fire  to  be  heated;  but  the  greater  part  of  the  heat  goes  off  in 
the  gases  that  are  formed  by  the  union  of  the  fuel  with  the  air. 
It  is  to  save  this  heat  that  we  sometimes  see  steam  boilers  set  up 
in  connection  with  the  furnaces  of  large  forge  shops. 

There  are  other  ways  of  generating  heat  besides  that  of  com- 
bustion. One  method,  that  is  coming  into  considerable  use  and 
which  is  especially  interesting  to  shop  men  because  of  the  ease 
with  which  it  can  be  controlled,  is  by  the  use  of  electricity.  We 
now  have  electric  annealing  and  hardening  furnaces  for  use  in 

16  163 


164  SHOP  ARITHMETIC 

tool  rooms,  where  a  close  regulation  of  the  heat  is  very  desirable. 
Then  there  are  the  electric  furnaces  by  which  aluminum  and 
carborundum  are  produced.  We  also  have  electric  welding  as 
an  example  of  the  production  of  heat  from  electricity. 

Another  method  of  heat  generation  that  is  frequently  encoun- 
tered in  shops,  often  where  it  is  not  desired,  is  the  production  of 
heat  from  work.  We  have  seen  how  heat  is  turned  into  work, 
but  here  we  have  work  returned  into  heat.  One  common  case 
of  this  is  in  bearings,  where  heat  is  produced  from  the  work  that 
is  spent  in  overcoming  the  friction.  Another  example  is  seen  in 
the  heating  of  a  lathe  tool  when  it  is  taking  a  heavy  cut,  or  in 
the  heating  of  the  tool  when  it  is  being  ground.  In  either  event, 
the  work  spent  in  removing  the  metal  goes  into  heat. 

123.  Temperatures. — Temperature  is  the  indication  of  the 
height  or  intensity  of  the  heat  in  a  body.  Lowering  the  tem- 
perature means  a  removal  of  heat  from  a  body,  and  raising  the 
temperature  means  the  addition  of  more  heat.  The  common 
method  of  measuring  temperature  is  by  means  of  an  instrument 
known  as  a  thermometer,  which  usually  consists  of  a  glass  tube 
which  is  partly  filled  with  mercury  and  which  has  the  air  ex- 
hausted from  the  other  part  of  it.  The  mercury  expands  and 
contracts  as  the  temperature  rises  or  falls  and,  therefore,  the 
height  of  the  column  of  mercury  is  a  measure  of  the  temperature. 
Alcohol  is  often  used  instead  of  mercury  for  outdoor  thermome- 
ters where  the  mercury  might  freeze. 

There  are  two  kinds  of  thermometer  scales  in  common  use — 
the  Centigrade  (abbreviated  C.)  and  the  Fahrenheit  (abbreviated 
Fahr.  or  F.) .  On  the  Centigrade  thermometer  the  space  between 
the  freezing-point  of  water  and  the  boiling-point  at  atmospheric 
pressure  is  divided  into  100  equal  parts  called  Degrees  (represented 
by  °)  the  freezing-point  being  marked  zero  (0°)  and  the  boiling- 
point  100°.  The  balance  of  the  scale  is  then  divided  into  spaces 
of  equal  length  below  zero  and  above  100°  in  order  that  tem- 
peratures higher  than  100  and  lower  than  zero  may  be  read. 

On  the  Fahrenheit  scale  the  freezing-point  of  water  is  marked 
32°  and  the  boiling-point  212°,  so  the  space  between  is  divided 
into  180°  (212° -32°  =  180°).  This  scale  is  also  marked  with 
divisions  below  32°  and  above  212°  in  order  to  make  the  thermom- 
eter read  through  a  wider  range.  The  Fahrenheit  thermometer 
is  used  more  commonly  in  the  United  States  than  the  Centigrade, 
which  is  used  extensively  in  Europe.  The  Centigrade  scale  is, 


HEAT 


165 


however,  used  in  this  country  for  most  scientific  work  and  is 
becoming  so  common  that  it  is  desirable  to  understand  the  rela- 
tions between  the  two  scales.  Fig.  82  shows  the  relation  of  the 
two  scales  up  to  212°  F.  or  100°  C. 

Since  the  same  interval  of  temperature  is  divided  into  100 
parts  in  the  Centigrade  scale,  and  180  parts  in  the  Fahrenheit 
scale,  each  Centigrade  degree  is  }£$,  or 
f  Fahrenheit  degrees.  Similarly,  one 
Fahrenheit  degree  is  f  of  a  Centigrade 
degree.  A  change  of  30°  in  tempera- 
ture on  the  Centigrade  scale  would 
equal  f  of  30,  or  54°  change  on  a 
Fahrenheit  thermometer.  Likewise, 
when  the  mercury  moves  27°  on  a 
Fahrenheit  thermometer,  it  would  move 
only  J-  of  27  =  15°  on  a  Centigrade  scale. 
In  changing  a  reading  on  one  thermom- 
eter scale  to  the  corresponding  reading 
on  the  other,  it  is  necessary  to  remem- 
ber that  the  zero  points  are  not  the 
same.  The  Centigrade  zero  is  at  32°  F. 
In  other  words,  the  two  zeros  are  32 
Fahrenheit  degrees  apart. 

To  change  a  reading  on  the  Centigrade 
scale  to  the  corresponding  Fahrenheit 
reading:  First  multiply  the  degrees  C. 
by  f .  This  gives  an  equivalent  number 
of  degrees  on  the  F.  scale.  To  this  add 
32,  in  order  to  have  the  reading  from 
the  F.  zero. 

To  change  a  reading  on  the  Fahrenheit 
scale  to  the  corresponding  Centigrade 
reading:  First  subtract  32.  This  gives  the  number  of  F.  de- 
grees above  freezing  (which  is  the  C.  zero) .  Multiply  the  result 
by  f ,  thus  obtaining  the  desired  C.  reading. 

Examples : 

1.  Change  30°  C.  to  the  corresponding  Fahrenheit  reading. 

30  X  r  =  54,  the  equivalent  number  of  F.  degrees. 

54+32-86°  F.,  the  reading  on  a  F.  thermometer. 

2.  What  would  a  Centigrade  thermometer  read  when  a  Fahrenheit  ther- 
mometer stood  at  72°? 


210 

200— 

190— 

180— 

170— 

160— 

150— 

140— 

130— 

120— 

110— 

100— 

so- 
so— 

70— 
60— 
BO— 
40  — 
F^0= 
20 — 
10— 
0— 
-10— 
-20— 
-30— 
-40— 


1 

212°F  =  100°C 

BOILING          i 

100  0 

-  —   90 

-  —   80 

-—    70 

60 

•  —    50 

1 

40 

| 

•  —   30 

- 

-—  20 

; 

•  —   10 

32°  F.  0°C        1 

• 

FREEZING        • 

•  —     o  C 

\ 

10 

\ 

20 

1 

30 

- 

1G6  SHOP  ARITHMETIC 

72  —  32  =  40,  the  number  of  F.  degrees  above  freezing. 

5         2° 
40  X  §  =  22—  C.,  Answer. 

These  rules  or  relations  are  often  expressed  by  the  following 
formulas,  in  which  C  stands  for  a  reading  on  the  Centigrade 
thermometer  and  F  for  a  reading  on  the  Fahrenheit  thermometer. 


The  parenthesis  (  )  when  used  as  above  means  that  the  work 
indicated  inside  of  it  is  to  be  done  first  and  then  the  result 
multiplied  by  f  .  In  the  second  formula,  the  C  is  first  to  be 
multiplied  by  f  and  then  the  32  is  added  to  the  product.  It  is 
always  to  be  understood  that  multiplications  and  divisions  are 
to  be  performed  before  additions  and  subtractions  unless  the 
reverse  is  indicated,  as  was  done  in  the  first  of  these  formulas,  by 
the  use  of  the  parenthesis  (  )  . 

When  it  comes  to  measuring  the  temperatures  in  furnaces,  as 
is  often  desirable  in  fine  tool  work,  a  thermometer  is  clearly  out 
of  the  question.  As  the  mercury  thermometer  is  ordinarily 
made,  it  should  not  be  used  for  temperatures  above  500°,  but, 
by  filling  the  glass  tube  above  the  mercury  with  nitrogen  gas 
under  pressure,  a  thermometer  can  be  made  that  may  be  read 
up  to  900°.  For  higher  temperatures,  devices  called  Pyrometers 
are  used.  There  are  numerous  kinds  of  pyrometers,  but  the  one 
most  used  in  the  shops  for  furnace  temperatures  is  what  is  called 
the  Le  Chatelier  pyrometer.  In  this  pyrometer,  one  end  of  a 
porcelain  tube  about  f  in.  in  diameter  and  from  12  in.  to  40  in. 
long  is  thrust  into  the  furnace  and  held  there  or,  if  frequent 
readings  are  to  be  taken,  it  may  be  placed  there  permanently. 
Inside  this  tube  are  some  wires  of  special  composition  that 
generate  an  electric  current  when  they  get  hot.  From  the  other 
end  of  the  tube  a  couple  of  wires  run  to  a  small  box  containing 
a  "galvanometer,"  that  is,  a  device  for  indicating  the  strength  of 
the  electric  current  generated.  This  has  a  needle  swinging  over 
a  dial  and  the  dial  is  usually  laid  off  in  degrees  so  the  temperature 
is  read  direct.  Most  of  these  pyrometers  have  centigrade  gradu- 
ations, but  one  should  be  sure  which  scale  a  pyrometer  has 
before  he  uses  it. 


HEAT  167 

For  example,  suppose  we  wanted  to  get  1000°  F.  and  by  mis- 
take had  1000°  C.  instead. 

?  X  1000  +  32=  1832°  F. 
o 

1000°  C.    =  1832°  F. 

This  shows  that  it  would  be  pretty  serious  to  use  the  wrrong  scale. 

It  has  for  years  been  the  practice  of  the  older  shop  men  to  tell 
the  temperature  of  steel  or  iron  by  its  color.  This  method  has 
its  disadvantages,  however,  as  so  much  depends  on  the  sensitive- 
ness of  the  man's  eye  and  on  whether  the  work  is  being  done  in 
bright  sunlight  or  in  a  dark  corner  of  the  shop.  A  bar  will  show 
red  in  the  dark  when  it  would  still  be  black  in  the  sunlight. 

For  the  lower  range  of  temperatures  (those  used  in  tempering 
tools)  we  can  judge  the  temperature  by  the  color  which  will 
appear  on  a  polished  steel  surface  when  heated  in  the  air.  These 
tempering  colors  and  their  uses  for  carbon  tool  steels  are  about 
as  follows: 

430°  F.     Very  pale  yellow  Scrapers 

Hammer  faces 

Lathe,  shaper,  and  planer  tools 
460°          Straw  yellow  Milling  cutters 

Taps  and  dies 
480°          Dark  straw  color  Punches  and  dies 

Knives 

Reamers 
f>00°          Brownish-yellow  Stone  cutting  tools 

Twist  drills 

.">20°          Yellow  tinged  with  purple     Drift  pins 
530°          Light  purple  Augers 

Cold  chisels  for  steel 
oo()°          Dark  purple  Hatchets 

Cold  chisels  for  iron 

Screw  drivers 

Springs 

570°          Dark  blue  Saws  for  wood 

610°         Pale  blue 
630°          Blue  tinged  with  green 

More  uniform  results  can  be  obtained  if  the  steel  is  heated  in 
a  bath  of  sand  or  of  oil,  the  bath  being  maintained  at  the 
desired  temperature  and  a  pyrometer  being  used  to  observe  the 
temperature.  For  higher  temperatures,  molten  lead  or  mineral 


1G8  SHOP  ARITHMETIC 

salts  such  as  common  salt,  barium  chloride,  potassium  chloride, 
and  potassium  cyanide  are  used. 

When  steel  and -iron  are  heated  to  higher  temperatures,  they 
successively  become  red,  orange,  and  white.  These  colors  and 
the  corresponding  temperatures  are  about  as  follows: 

957°  F.  First  signs  of  red 

1290°  Dull  red 

1470°  Dark  cherry 

1655°  Cherry  red 

1830°  Bright  cherry 

2010°  Dull  orange 

2190°  Bright  orange 

2370°  White  heat 

2550°  Bright  white — welding  heat 
2730°       ^ 

2910°  t0  }  Dazzling  white. 

124.  Expansion  and  Contraction. — Nearly  all  substances 
expand  when  heat  is  applied  to  them  and  contract  when  heat  is 
removed.  This  phenomenon  is  greatest  in  gases  and  least  in 
solids,  but  even  in  solids  it  is  of  enough  moment  to  be  extremely 
useful  at  times  or  to  cause  considerable  trouble  when  allowance 
is  not  made  for  it. 

There  are  a  few  metal  alloys  which,  within  certain  limits,  do 
not  change  their  volumes  with  changes  of  temperature,  and  there 
are  also  some  which  between  certain  temperatures  will  even 
expand  when  cooled  and  contract  when  heated.  A  nickel  steel 
containing  36%  nickel  has  practically  no  expansion  or  contrac- 
tion with  changes  in  temperature  and  is,  therefore,  used  in  some 
cases  for  accurate  measurements  where  expansion  of  the  measur- 
ing instruments  would  introduce  serious  errors. 

When  a  solid  body  is  heated  it  expands  in  all  directions,  if 
free  to  do  so,  but  as  a  rule  we  are  concerned  only  with  the  change 
of  one  dimension  and  not  with  the  change  in  volume.  Thus,  in 
the  case  of  a  steam  pipe  we  do  not  care  about  the  change  in 
thickness  or  in  diameter,  but  we  are  concerned  with  the  change  in 
length.  On  the  other  hand,  when  a  bearing  gets  hot  and  seizes, 
it  is  the  change  in  diameter  that  causes  the  trouble.  There  are 
few  machinists  who  have  not  had  the  experience,  in  boring  a 
sleeve  to  fit  a  certain  shaft,  of  having  a  free  fit  when  tested  just 
after  taking  a  cut  through  the  sleeve,  and  then  later  of  finding 
that  the  sleeve  fitted  so  tightly  that  it  had  to  be  driven  off  the 
shaft.  Of  course,  the  explanation  is  that  the  sleeve  becomes 


HEAT  169 

warm  when  being  bored  in  the  lathe,  while  the  shaft  is  much 
cooler.  When  the  sleeve  cools  to  the  temperature  of  the  .shaft,  it 
contracts  and  seizes  or  "freezes"  to  the  shaft.  In  accurate  tool 
work  the  effect  of  differences  in  temperature  between  the  measur- 
ing instruments  and  the  work  may  become  serious.  For  this 
reason  many  gages  are  provided  with  rubber  or  wooden  handles 
which  do  not  conduct  heat  readily.  They  thus  prevent  the  heat 
of  the  hand  from  getting  into  the  gages  and  expanding  them. 

But  this  is  enough  to  give  some  idea  of  the  troubles  caused  by 
this  property  of  materials;  let  us  now  see  of  what  benefit  it  is. 
We  have  already  seen  the  use  that  is  made  of  the  expansion  of 
mercury  in  thermometers.  There  are  numerous  heat  regulating 
devices"  (called  thermostats)  which  depend  on  the  expansion 
or  contraction  of  a  bar  to  perform  the  desired  operations.  We 
find  these  used  for  regulating  house  heating  boilers  and  furnaces, 
incubators,  anti.  other  devices  where  uniform  temperatures  are 
required.  Probably  the  greatest  shop  use  of  expansion  and 
contraction  is  in  making  shrink  fits.  When  we  want  to  fasten 
securely  and  permanently  one  piece  of  metal  around  another,  we 
generally  shrink  the  first  onto  the  second.  This  process  is  used 
for  attaching  all  sorts  of  bands  and  collars  to  shafts,  cylinders, 
and  the  like,  for  putting  tires  on  locomotive  wheels,  and  for 
similar  work.  The  erecting  engineer  uses  it  to  put  in  the  links  in 
a  sectional  fly-wheel  rim  or  to  draw  up  bolts  in  the  hub  or  in  any 
other  place  where  he  wants  to  make  a  rigid  permanent  joint. 

The  amount  of  linear  expansion  which  a  body  undergoes 
depends  upon  the  kind  of  material  of  which  the  body  is  made, 
upon  the  amount  of  the  temperature  change  and,  of  course, 
upon  the  original  length. 

The  coefficient  of  linear  expansion  of  a  substance  is  that  part 
of  its  original  length  which  a  body  will  expand  for  each  degree 
change  in  temperature.  Coefficients  for  different  metals  have 
been  determined  for  our  use  by  careful  experiments,  and  can 
be  found  in  hand  books  or  tables  under  the  head  of  "Coefficients 
of  Expansion."  The  values  given  in  different  books  do  not  al- 
ways agree.  In  fact,  the  exact  compositions  of  the  metals  used 
in  the  tests  were  undoubtedly  different  for  the  different  tests  that 
are  on  record.  Hence,  different  tables  give  slightly  different 
rates  of  expansion.  The  following  values  are  taken  from  the 
most  reliable  authorities  and  are  sufficiently  accurate  for  most 
purposes. 


170 


SHOP  ARITHMETIC 
COEFFICIENTS  OF  EXPANSION 


Metal 

Coefficient 

Aluminum  

.  00001234 

Brass  

.  00001 

Cast  iron  

.0000055  to 

Wrought  iron  and  machine  steel  . 
36%  nickel  steel  

.000006 
.0000065 
.  0000003 

The  above  values  are  based  on  a  temperature  rise  of  1°  F. 
For  one  Centigrade  degree  change  in  temperature  the  coefficients 
would  be  f  of  those  just  given.  The  student  is  not  expected  to 
memorize  these  values.  Remember  that  if  the  length  is  given 
in  feet  the  expansion  calculated  will  be  in  feet,  and  if  the  length  is 
in  inches  the  expansion  calculated  will  be  in  inches.  To  get 
the  actual  expansion  per  degree  for  any  certain  length,  multiply 
the  coefficient  of  expansion  by  the  length.  If  the  temperature 
change  is  100°,  the  expansion  will  be  100  times  that  for  1°. 

Example  : 

The  head  of  a  gas  engine  piston  in  operation  has  a  temperature  of 
about  400°  higher  than  the  cylinder  in  which  it  is  running.  What  allowance 
must  be  made  for  this  expansion  in  a  12  in.  piston?  (The  piston  is  made  of 
cast  iron.) 

.000006X400  =  .0024  in.  expansion  per  inch 
.0024X12        =.0288  in.  expansion  in  12  in.,  Answer. 

The  head  of  the  piston  must,  therefore,  be  turned  at  least  .0288  in.  small  to 
allow  for  the  expansion  to  take  place  without  the  piston  seizing  in  the 
cylinder. 

The  law  of  expansion  and  contraction  may  be  expressed  by  a 
formula  as  follows: 


where 


E =  TxCxL 

E  is  the  change  in  length 

T  is  the  change  in  temperature 

C  is  the  coefficient  of  linear  expansion 

L  is  the  original  length  of  the  body. 


HEAT  171 

Example : 

What  will  be  the  expansion  in  a  steam  pipe  200  ft.  long  when 
subjected  to  a  temperature  of  300°,  if  erected  when  the  temperature  was  60°? 

77  =  300-60  =  240;  C  =  .0000065;  L  =  200 
E=TXCXL 

=  240  X. 0000065X200  =  .312  ft.,  Answer. 

Notice  particularly  that  here  we  use  L  in  feet  and,  consequently,  the  expan- 
sion E  comes  out  in  feet.  This  can  be  reduced  to  inches  if  desired,  giving 
12 X. 312  =  3. 744  in.  or  3|  in.  nearly. 

125.  Allowances  for  Shrink  Fits. — In  making  a  shrink  fit,  the 
collar  or  band,  or  whatever  is  to  be  shrunk  on,  is  bored  slightly 
smaller  than  the  outside  diameter  of  the  part  on  which  it  is  to 
be  shrunk.  It  is  then  heated  and  thus  expanded  until  it  can  be 
slipped  into  place.  When  it  cools,  it  cannot  return  to  its  original 
size  but  is  in  a  stretched  condition.  It,  therefore,  exerts  a  power- 
ful grip  on  the  article  over  which  is  has  been  shrunk. 

Practice  differs  considerably  in  the  allowances  that  are  made 
for  shrink  fits.  A  rule  which  has  been  widely  and  successfully 
used  is  to  allow  oVu"  m<  f°r  each  inch  of  diameter.  According 
to  this  rule,  if  we  were  shrinking  a  crank  on  a  6  in.  shaft,  the 
crank  should  be  bored  .  006  in.  small  or  else  the  shaft  turned  .  006 
in.  oversize  and  the  crank  bored  exactly  6  in.  For  a  10-in.  shaft 
we  would  allow  .010  in,  and  so  on  for  other  sizes. 

This  could  be  expressed  by  the  following  formulas: 

A  =  .OOlXl>,  or,  since  -001=77)7)7),  this  could  be  written 
D 


1000 

where  A  stands  for  "  allowance  " 
and     D  for  the  diameter 

Assuming  that  an  allowance  of  .001 XD  is  made,  let  us  see 
what  temperature  is  necessary  in  order  to  give  the  necessary 
expansion  so  that  a  steel  tire  can  be  put  over  a  locomotive  driving 
wheel. 

For  each  degree  that  the  tire  is  heated,  it  will  expand  .0000065 
in.  per  inch  of  diameter.  We  must  have  an  expansion  of  at  least 
.001  in.  The  number  of  de'grees  necessary  to  get  this  will  be 

.001 


. 0000065 

17 


=  154 


172  SHOP  ARITHMETIC 

It  would  look  as  if  a  difference  of  154°  would  be  sufficient. 
However,  a  greater  difference  is  necessary  in  practice.  There 
must  be  sufficient  clearance  so  that  the  tire  can  be  slipped  quickly 
into  place  before  it  has  time  to  cool  off  or  to  warm  the  wheel. 
Once  in  place,  the  tire  will  grip  the  wheel  when  a  temperaUire 
difference  of  154°  exists. 

PROBLEMS 

211.  In  testing  direct  current  generators,  it  is  customary  to  specify  that 
under  full  load  the  temperature  of  the  armature  shall  not  rise  more  than  40° 
Centigrade  above  a  room  temperature  of  25°  C. ;  that  is,  the  temperature 
of  the  armature  under  these  conditions  should  not  exceed  65°  C. 

In  making  a  test  a  Fahrenheit  thermometer  was  used.  The  room  was  at 
a  temperature  of  77°  F.  and  the  temperature  of  the  armature  at  the  end  of 
the  run  was  180°  F.  Did  the  generator  meet  the  specifications?  What 
was  the  temperature  change,  Centigrade? 

212.  In  erecting  a  long  steam  line  that  will  have  a  variation  in  tempera- 
ture of  320°,  how  far  apart  should  the  expansion  joints  be  placed  if  each 
joint  can  take  care  of  a  motion  of  3  inches? 

213.  If  a  brass  bushing  measures  2  in.  just  after  boring,  when  its  tem- 
perature is  95°  F.,  what  will  it  caliper  when  it  has  cooled  to  65°  F.? 

214.  A  steel  link  2  ft.  long  is  made  ^  in.  too  short  for  the  slot  in  the  fly- 
wheel rim  into  which  it  is  to  be  shrunk.     How  hot  must  the  link  be  before 
it  will  go  in? 

215.  If  a  hub  bolt  is  heated  until  it  just  begins  to  show  red  and  is  im- 
mediately screwed  up  snug  and  allowed  to  cool,  what  shrinkage  allowance 
per  inch  of  length  would  we  be  allowing  by  such  a  plan? 

216.  If  we  wished  to  maintain  a  tempering  bath  at  a  temperature  of 
500°  F.,  what  should  be  the  reading  on  a  Centigrade  pyrometer? 

217.  If  the  brass  bearings  for  a  2  in.  steel  crank  shaft  are  given  a  running 
clearance  of  .002  in.  at  a  temperature  of  60°  F.,  what  would  be  the  clearance 
when  running  at  a  temperature  of  100°  F.? 

218.  A  horizontal  steam  turbine  and  dynamo  are  to  be  direct-connected, 
their  shaft  centers  being  3£  ft.  above  the  bed  plate.     If  the  bearings  are 
lined  up  at  a  temperature  of  70°,  how  much  will  they  be  out  of  alignment 
under  running  conditions  when  the  temperature  of  the  dynamo  frame  is 
80°  F.  and  that  of  the  turbine  is  215°  F.,  both  frames  being  of  cast  iron? 


CHAPTER  XX 
STRENGTH  OF  MATERIALS 

126.  Stresses. — When  a  load  is  put  upon  any  piece  of  material, 
it  tends  to  change  the  shape  of  the  piece.  The  material  naturally 
resists  this  and,  therefore,  exerts  a  force  opposite  to  the  load.  If 
the  load  is  not  too  heavy,  the  material  may  be  able  to  exert  a 
sufficient  force  to  hold  it,  but  often  the  strength  of  the  material 
is  exceeded  and  the  piece  breaks. 

The  resistance  which  is  set  up  when  a  piece  of  material  is  loaded 
is  called  the  Stress.  For  instance,  if  a  casting  weighing  3  tons  or 
6000  Ib.  is  suspended  by  a  single  rope,  the  stress  in  the  rope  will 
be  6000  Ib. 

x\\\\\\ \v\\\\ 


W 


tension 


\\\\\\\\\\\VC 

compression 
FIG.  83. 


shear 


There  are  three  different  kinds  of  stresses  that  can  be  produced, 
depending  on  the  way  the  load  is  applied 

(1)  Tensile  stress  (pulling  stress). 

(2)  Compressive  stress  (crushing  or  pushing  stress) . 

(3)  Shearing  stress  (cutting  stress). 

Fig.  83  shows  how  these  different  stresses  are  produced. 
We  sometimes  recognize  two  other  kinds  of  stresses,  but  these 
are  really  special  cases  of  the  three  just  given.     These  two  others 


are: 


173 


174 


SHOP  ARITHMETIC 


(a)  Bending  Stress  (really  a  combination  of  tension  on  one 
side  and  compression  on  the  other). 

(b)  Torsional  or  Twisting  Stress  (a  form  of  shearing  stress) . 
127.  Ultimate  Strengths. — By  taking  specimens  of  the  different 

materials  and  loading  them  until  they  break,  it  has  been  possible 
to  find  out  just  what  each  kind  of  material  will  stand.  The  load 
to  which  each  square  inch  of  cross-section  must  be  subjected  in 
order  to  break  it,  is  called  the  Ultimate  strength  of  the  material. 
The  strength  of  most  materials  differs  for  the  different  methods 
of  loading  shown  in  Fig.  83. 

The  Tensile  Strength  of  a  material  is  the  resistance  offered  by 
its  fibers  to  being  pulled  apart. 

The  Compressive  Strength  of  a  material  is  the  resistance 
offered  by  its  fibers  to  being  crushed. 

The  Shearing  Strength  of  a  material  is  the  resistance  offered 
by  its  fibers  to  being  cut  off. 

The  following  table  gives  the  average  values  for  the  most  used 
materials. 

ULTIMATE  STRENGTHS— POUNDS  PER  SQUARE  INCH 


Material 

Tension 

Compression 

Shear 

Timber 

10000 

8000 

3000  (across  grain) 

Cast  iron   

20000 

90000 

20000 

Wrought  iron  
Machine  steel  

50000 
65000 

50000 
65000 

40000 
50000 

128.  Safe  Working  Stresses. — Having  found  how  great  a  stress 
is  required  to  break  one  square  inch  of  material,  we  naturally 
would  not  allow  anywhere  near  this  stress  to  come  on  a  piece 
of  material  in  actual  service.  The  Ultimate  Strength  is  usually 
divided  by  some  number,  known  as  the  Factor  of  Safety,  and  the 
quotient  is  used  as  the  Safe  Working  Stress. 

For  example,  if  60,000  Ib.  per  square  inch  will  break  a  piece  of 
soft  steel  and  we  use  a  factor  of  safety  of  5,  this  would  give: 

(\C\OC\O 

Safe  working  stress  =  — = —  =  12000   Ib.    per   square  inch. 

o 

The  following  table  gives  the  Safe  Working  Stresses  of  the  most 
used  materials. 


STRENGTH  OF  MATERIALS  175 

SAFE  WORKING  STRESSES— POUNDS  PER  SQUARE  INCH 


Material 

8t 

Tension 

Sc 
Compression 

s, 

Shear 

Timber  

700 

700 

500 

Cast  iron  .... 

3-  4000 

15-18000 

3-  4000 

Wrought  iron  .... 
Machine  steel.  .  .  . 

8-10000 
10-16000 

8-10000 
10-16000 

7-  9000 
8--12000 

Instead  of  writing  "  safe  loads  in  pounds  per  square  inch  "  for 
tension,  compression,  or  shear,  the  symbols  St,  Sc,  and  S8  are  used. 
So  if  A  =  area  in  square  inches,  then  the  load  W  which  can  be 
carried  safely  =  Area  X  safe  load  per  square  inch  or 
AxSt  =  W  (tension) 
A  X  Sc  =  W  (compression) 
AxSa  =  W  (shear) 
Or,  in  general,  for  all  stresses 

AXS  =  W 

Perhaps  more  often  we  would  want  to  find  the  area  necessary  in 
order  to  support  a  certain  weight  or  load.  In  this  case,  we  would 
want  a  formula  which  would  give  A. 

If  we  divide  the  total  load  by  the  safe  stress,  we  will  get  the 
necessary  area;  or 

W 


A  = 


S 


This  simply  says,  area  of  metal  necessary  =  total  weight  to  be 
carried  divided  by  safe  load  in  pounds  per  square  inch.  From 
the  area  of  a  bolt  or  rod,  its  diameter  can  be  easily  found. 

129.  Strengths  of  Bolts. — There  is  a  well-known  saying  that 
"  a  chain  is  only  as  strong  as  its  weakest  link."  This  means,  in 
general,  that  any  mechanism  must  be  so  designed  that  its  weakest 
part  will  be  strong  enough  to  stand  the  greatest  load  that  may 
come  on  it.  In  figuring  the  size  of  a  bolt  to  hold  a  certain  load, 
we  would  not  calculate  the  full  diameter  of  the  bolt  and  make 
the  area  there  just  sufficient,  but  we  must  see  to  it  that  the  bolt 
has  a  cross-sectional  area  at  the  root  of  the  threads  large  enough  to 
support  the  load.  Then  the  body  of  the  bolt  will  have  a  surplus 
of  strength. 


176 


SHOP  ARITHMETIC 


Example : 

'  What  size  of  steel  eyebolt  will  support  a  weight  of  5000  lb.? 
Take  12,000  lb.  as  the  safe  load  in  tension. 

W       5000 

Then'A=T=l2ooo 

5 
A  =   2  sq.  in.  =  .416  sq.  in. 

.416  sq.  in.  is  then  the  necessary  area  to  support  the  weight.  Of  course, 
the  example  could  be  completed  by  saying  .7854  D2  =  .416  sq.  in.,  where 
D  =  diameter  at  the  root  of  the  thread.  By  then  solving  for  D  we  would  get 
the  diameter  at  the  root  of  the  threads.  But  the  Bolt  Tables  afford  an 
easier  method  than  this.  In  the  following  table,  .4193  is  given  as  the  area 
of  a  |  in.  bolt  at  the  root  of  the  thread.  Therefore,  a  |  in.  eyebolt  would 
probably  be  used. 

In  figuring  the  allowable  loads  for  steel  bolts,  it  is  best  not 
to  allow  over  12,000  lb.  stress  per  square  inch  and  10,000  lb.  is 
perhaps  even  more  usual  on  account  of  the  sharp  root  of  the 
threads,  which  makes  a  bolt  liable  to  develop  cracks  at  this  point. 

BOLT  TABLE.— U.  S.  S.  THREADS 


Diam. 

Threads 
to  inch 

Diam.  at 
bottom  of 
thread 

Area  of  bolt 

Area  at 
bottom  of 
thread 

tin. 

20 

.1850 

.0491 

.0269 

•fs  in. 

18 

.2403 

.0767 

.0454 

1  in. 

16 

.2938 

.1104 

.0678 

T7T  in. 

14 

.3447 

.1503 

.0933 

i  in. 

13 

.4001 

.1963 

.1257 

-,9,T  in. 

12 

.4542 

.2485 

.1621 

fin. 

11 

.5069 

.3068 

.2018 

J  in. 

10 

.6201 

.4418 

.3020 

I  in- 

9 

.7307 

.6013 

.4193 

1  in. 

8 

.8376 

.7854 

.5510 

IJin. 

7 

.9394 

.9940 

.6931 

Hin. 

7 

1  .  0644 

1  .  2272 

.8899 

If  in. 

6 

1.1585 

1.4849 

1.0541 

liin. 

6 

1.2835                1.7671                1.2938 

If  in. 

5i 

1.3888                2.0739                1,5149 

If  in. 

5 

1.4902                2.4053                1.7441 

2  in. 

4i 

1.7113                3.1416                2.3001 

2\  in. 

4i 

1.9613                3.9761                3.0213 

2iin. 

4 

2.1752                4.9087                3.7163 

2|  in. 

4 

2.4252                5.9396                4.6196 

3  in. 

34 

2.6288                7.0686                5.4277 

STRENGTH  OF  MATERIALS  177 

130.  Strength  of  Hemp  Ropes.  —  It  is  quite  common  in  calcu- 
lating the  strength  of  ropes  and  cables  to  assume  that  the  section 
of  the  rope  is  a  solid  circle.  Of  course,  the  strands  of  the  rope 
do  not  completely  fill  the  circle  but,  if  we  find  by  test  the  allow- 
able safe  strength  per  square  inch  on  this  basis,  it  will  be  perfectly 
safe  to  make  calculations  for  other  sizes  of  ropes  on  the  same 
basis.  The  safe  working  stress  based  on  the  full  area  of  the 
circle  is  1420  Ib.  per  square  inch.  The  Nominal  Area  (as  the 
area  of  the  full  circle  by  which  the  rope  is  designated  is  called)  is 
A  =  .7854X.D2.  The  safe  stress  is  1420  Ib.  per  square  inch  and, 
consequently,  the  weight  that  can  be  supported  by  a  rope  of 
diameter  D  is 

W  =  SxA 

=  1420  X.  7854  X#2 

Here  we  have  two  constant  numbers  (1420  and  .7854)  that 
would  be  used  every  time  we  were  to  calculate  the  safe  strength 
of  a  rope.  If  this  were  to  be  done  often  we  would  not  want  to 
multiply  these  together  every  time,  so  we  can  combine  them  now, 
once  and  for  all. 

1420  X.  7854  =  1120,  approximately 

Hence 

*    TF  =  1120XZ)2 

Example  : 

Find  the  safe  load  on  a  hemp  rope  of  £  in.  diameter. 


=  1120X4  =  280  Ib.,  Answer. 

131.  Wire  Ropes  and  Cables.  —  For  wire  ropes  made  of  crucible 
steel,  a  safe  working  load  of  15,000  Ib.  per  square  inch  of  nominal 
area  is  allowable.     For  cables  of  Swedish  iron  but  half  this  value 
should  be  used. 

132.  Strength  of  Chains.  —  It  has  been  demonstrated  by  re- 
peated tests  that  a  welded  joint  cannot  be  safely  loaded  as  heavily 
as  a  solid  piece  of  material.     Of  course,  there  are  often  welds 
that  are  practically  as  strong  as  the  stock,  but  it  is  not  safe  to 
depend  on  them.     For  this  reason,  the  safe  working  load  per 


178  SHOP  ARITHMETIC 

square  inch  for  chain  links  is  often  given  as  9000  lb.,  which  is 
just  f  of  12,000  lb. 

If  D  =  the  diameter  of  the  rod  of  which  the  links  are  made 
A=2X.7854X£>2 
W  =  StXA 
F  =  9000X2X.7854X£>2 

Combining  the  constant  numbers,  this  can  be  simplified  into 

W  =  14,000  X£>2 
This  is  used  in  the  same  way  as  the  formula  for  a  rope. 

133.  Columns. — The  previous  examples  were  cases  of  tension. 
The  size  of  a  rod  or  timber  subjected  to  compression  is  computed 
in  the  same  way  unless  it  is  long  in  comparison  with  its  thickness. 
When  a  bar  under  compression  has  a  length  greater  than  ten 
times  its  least  thickness,  it  is  called  a  Column  and  must  be  con- 
sidered by  the  use  of  complicated  formulas  which  take  account 
of  its  length.  It  can  be  seen  by  taking  a  yardstick,  or  similar 
piece,  that  it  is  much  easier  to  break  than  a  piece  of  shorter 
length  but  otherwise  of  the  same  dimensions.  A  long  piece, 
when  compressed,  will  buckle  in  the  center  and  break  under  a 
light  thrust  or  compression.  An  example  of  this  can  be  found 
in  the  piston  rod  on  a  steam  engine,  where,  on  account  of  the 
length  of  the  rod,  it  is  necessary  to  use  much  lower  stresses  than 
those  given  in  the  tables.  The  compressive  stress  allowed  in 
piston  rods  varies  with  the  judgment  of  different  designers  but 
is  generally  about  5000  lb.  per  square  inch,  using  a  pressure  on 
the  piston  of  125  lb.  per  square  inch. 

Example : 

Find  the  size  of  rod  for  a  30  in.  by  52  in.  Corliss  engine  with  125  lb . 
steam  pressure. 

30  in.  is  the  diameter  of  the  cylinder  and  52  in.  is  the  stroke,  which  is  not 
considered  in  the  problem  except  in  that  it  has  reduced  the  allowable  stress 
in  the  rod. 

.7854  X302  =  706.86  sq.  in.,  area  of  piston. 
706.86X125  =  88357.5  lb.,  total  pressure  on  piston. 
Using  5000  lb.  per  square  inch,  allowable  stress  in  the  rod. 

88358-7-5000  =  17.67  sq.  in.  sectional  area  of  rod, 

From  the  table  of  areas  of  circles,  it  is  seen  that  this  is  the  area  of  a  circle 
nearly  4f  in.  in  diameter,  so  we  would  use  a  4£  in.  rod. 

PROBLEMS 

Note. — In  all  examples  involving  screw  threads,  to  get  areas  at  root  of 
thread,  use  the  table  given  in  this  chapter.  Give  sizes  of  bolts  always  as 
diameters. 


STRENGTH  OF  MATERIALS 


179 


219.  If  the  generator  frame  shown  in  Fig.  84  weighs  3000  lb.,  what  size 
steel  eyebolt  should  be  used  for  lifting  it,  allowing  a  stress  of  10,000  lb.  at 
the  root  of  the  thread? 

220.  What  would  be  the  safe  load  for  a  J  in.  chain? 

221.  What  size  hemp  rope  would  be  necessary  to  lift  a  load  of  4000  lb.? 


Fio.  84. 

222.  What  force  would  be  necessary  to  shear  off  a  bar  of  machinery 
steel  2  in.  in  diameter? 

223.  A  certain  manufacturer  of  jack  screws  states  that  a  2J  in.  screw  is 
capable  of  raising  28  tons.     If  the  diameter  of  the  screw  at  the  base  of  the 
threads  is  1.82  in.,  what  is  the  stress  per  square  inch  at  the  bottom  of  the 
threads  when  carrying  28  tons? 

224.  A  soft  steel  test  bar  having  a  diameter  of  .8  in.  is  pulled  in  two  by  a 
load  of  31,500  lb.     What  was  the  breaking  tensile  stress  per  square  inch? 


Fio.  85. 


225.  The  cylinder  head  of  a  small  steam  engine  (Fig.  85)  having  a  cylinder 
diameter  of  7  in.  is  held  on  by  6  studs  of  $  in.  diameter.     When  there  is  a 
steam  pressure  of  125  lb.  per  square  inch  in  the  cylinder,  what  will  be  the 
pull  on  each  stud?     And  what  will  be  the  stress  per  square  inch  in  each  stud, 
due  to  the  steam  pressure? 

226.  With  a  cylinder  diameter  of  10  in.  and  an  air  pressure  of  100  lb. 
per  square  inch,  find  the  greatest  weight  that  can  be  lifted  by  the  air  hoist, 


180 


SHOP  ARITHMETIC 


shown  in  Fig.  86.  Also  find  the  size  of  piston  rod  necessary,  assuming  that 
it  is  screwed  into  the  piston.  Notice  that  this  rod  is  subject  only  to  tension 
and,  therefore,  a  greater  stress  is  allowable  than  in  steam  engine  piston  rods. 

227.  Work  out  a  formula  for  the  strength  of  crucible  steel  cables  on  the 
same  plan  as  that  given  for  hemp  rope. 

228.  What  is  the  greatest  load  that  should  be  lifted  with  a  pair  of  tackle 
blocks  having  3  pulley  in  the  movable  block  and  2  in  the  fixed  block,  and 
having  a  f  in.  rope. 


m 


m 


FIG.  86. 


INDEX 

Addition  of  decimals,  35 

of  fractions,  1 1 
Air  compressors,  149 

steam  and,  159 
Allowances  for  shrink  fits,  171 
Analyzing  practical  problems  in  fractions,  22 
Area  of  a  circle,  77 
Areas  of  circles,  table  of,  101 
Arrangement  of  pulleys,  142 

Belting,  horse  power  of,  139 

rules  for,  141 
Belt  joints,  143 
Belts,  grain  and  flesh  sides,  143 

speeds  of,  141 

tension  per  inch  of  width,  141 

thickness  of,  141 

width  of,  140 

Blocks,  types  of  tackle,  123 
Bolts,  strength  of,  175 
Bolt  table,  U.  S.  S.  threads,  170 

Cables,  strength  of  wire  rope  and,  177 
Cancellation,  19 
Casting,  weight  of,  83 
Centigrade  thermometers,  164 
Chains,  strength  of,  177 
Circle,  area  of,  77 

circumference  of,  51 

diameter  of,  51 

radius  of,  51 

Circumference  of  a  circle,  51 
Circumferences  of  circles,  table  of,  101 
Circumferential  speeds,  54 
Classes  of  levers,  117 
Coefficient  of  linear  expansion,  169 
Columns,  178 
Common  denominator,  9 

fractions  reduced  to  decimals,  39 
Complex  decimals,  40 
Compound  fractions,  21 

gear  and  pulley  trains,  68 
Contraction,  expansion  and,  168 

19  181 


182  INDEX 

Cube,  the,  80 

root,  75,  92 

roots  of  decimals,  98 

of  numbers  greater  than  1000,  97 

table,  103 
Cubes,  75 

and  higher  powers,  75 

of  numbers,  table  of,  103 
Cubical  measure,  units  of,  80 
Cutting  speeds,  57 

Decimal  equivalents,  table  of,  42 

fractions,  33 
Decimals,  addition  of,  35 

complex,  40 

cube  roots  of,  98 

division  of,  37 

multiplication  of,  36 

short  cuts,  37 

subtraction  of,  35 
Denominator,  common,  9 

least  common,  10 

of  a  fraction,  2 
Diameter  of  a  circle,  51 

from  area,  calculation  of,  91 
Differential  pulleys,  126 

pulley,  mechanical  advantage  of,  127 
Dimensions  of  circles,  91 

rectangles,  91 

squares,  91 

Direct  and  inverse  proportions,  65 
Division  of  decimals,  37 

fractions,  21 

Efficiencies,  134 

Efficiency  of  engines,  mechanical,  151 

of  hydraulic  jack,  158 

of  jack  screw,  135 
Emery  wheels,  55 
Expansion  and  contraction,  168 
Extracting  square  root,  86 

cube  root,  96 

Fahrenheit  thermometers,  164 
Fluids,  153 

transmission  of  pressure  through,  154 
Foot  pound,  137 
Formulas,  52 


INDEX  183 


Fractions,  addition  of,  11 

common  fractions,  reducing  to  decimals,  39 

compound,  21 

decimal,  33 

definition  of,  2 

denominators  of,  2 

division  of,  21 

improper,  3 

multiplication  of,  17 

numerators  of,  2 

proper,  3 

reduction  of,  3 

subtraction  of,  12 

whole  number  times  a  fraction,  17 

writing  and  reading  of,  2 
Frictional  horse  power,  151 

Gas  engines,  147 

horse  power  of,  147 
Gear  ratios,  64 

Gears,  relation  of  sizes  and  speeds  of,  64 
Gear  trains,  66 

compound,  68 
Grindstones  and  emery  wheels,  55 

Heat,  nature  of,  163 
Horse  power,  138 

brake,  149 

frictional,  151 

of  belting,  139 

of  gas  engines,  147 

of  steam  engines,  145 
Hydraulic  heads,  158 

jack,  155 

mechanical  advantage  of,  156 

efficiency  of,  158 
Hypotenuse,  89 

Improper  fractions,  3 

reduction  of,  5 
Inclined  planes,  mechanical  advantage  of,  131 

theory  of',  130 

use  of,  130 
Interpolation,  96 
Inverse  proportion,  65 
Iron  and  steel,  colors  at  different  temperatures,  167 

Jack  screws,  133 

efficiency  of,  135 

mechanical  advantage  of,  134 


184  INDEX 

Law  of  right  triangles,  89 
Least  common  denominator,  10 
Levers,  115 

classes  of,  117 

compound,  118 

mechanical  advantage  of,  119 
Linear  expansion,  coefficient  of,  169 

Machines,  types  of,  115 
Mean  effective  pressure,  146 
Measure,  cubical,  80 

square,  76 
Measures  of  length,  7 

of  time,  7 

of  volume,  80 

Mechanical  advantage  of  differential  pulley,  127 
hydraulic  jack,  156 
inclined  plane,  131 
jack  screw,  134 
lever,  119 
tackle  blocks,  125 
wedge,  132 

efficiency  of  an  engine,  151 
Metals,  weights  of,  82 
Micrometer,  the,  40 
Mill,  the,  28 
Mixed  numbers,  3 

multiplication  of,  18 

reduction  of,  6 
Money,  U.  S.,  24 

addition,  25 

division,  26 

multiplication,  26 

reducing  cents  to  dollars,  28 

reducing  dollars  to  cents,  27 

subtraction,  26 

table  of,  28 

the  "mill,"  28 
Multiplication  of  decimals,  36 

fractions,  17 

mixed  numbers,  18 

whole  numbers  and  fractions,  17 

Nature  of  heat,  163 
Numbers,  mixed,  3 

multiplication  of,  18 
Numerator  of  a  fraction,  2 

Percentage,  44 


INDEX  185 


Percentage,  classes  of  problems  under,  48 

uses  of,  46 
Peripheral  speed,  54 
Periphery,  54 
Planes,  inclined,  130 
Plates,  short  rule  for  weights  of,  83 
Power,  138 

Powers,  cubes  and  higher,  75 
Pressure  through  fluids,  transmission  of,  154 
Proper  fractions,  3 
Proportion,  59 

direct  and  inverse,  65 
Pulleys  and  belts,  58 
Pulleys,  arrangement  of,  142 

diameters  of,  63 

distance  between  centers  of,  142 

speeds  of,  63 

Pulley  trains,  compound,  68 
Pyrometers,  166 

Radius  of  a  circle,  51 
Ratio  and  proportion,  59 
Rectangle,  the,  79 
Rectangles,  dimensions  of,  91 
Reduction  of  fractions,  3 

of  improper  fractions,  5 

of  mixed  numbers,  6 
Right  triangles,  89 
Rim  speed,  54 

Roots  of  numbers,  by  table,  square,  99 
Roots  of  numbers,  cube,  75,  92 

square,  75 

table  of,  103 
Ropes,  strengths  of,  177 
Rules  for  area  of  a  circle,  78 

belting,  141 

gears,  64 

pulleys,  63 

square  root,  88 

weights  of  plates,  83 

Safe  working  stresses,  174 
Screw  cutting,  72 
Shrink  fits,  allowances  for,  171 
Specific  gravity,  153 
Speeds  of  pulleys,  63 

circumferential,  54 

cutting,  57 

peripheral,  54 


186  INDEX 

Speeds,  rim,  54 
surface,  54 
Square  measure,  76 
table  of,  77 
root,  75 

by  table,  99 
extracting,  86 
meaning  of,  85 
rules  for,  88 

roots  of  numbers,  table  of,  103 
Squares  of  numbers,  table  of,  103 
Steam  and  air,  159 
engines,  145 

horse  power  of,  145 
Strengths  of  bolts,  175 

of  cables  and  wire  ropes,  177 
of  chains,  177 
of  hemp  ropes,  177 
of  wire  ropes  and  cables,  177 
Stresses,  definition  of  kinds  of,  173 

safe  working,  174 
Subtraction  of  decimals,  35 

of  fractions,  12 
Surface  speed,  54 

Tables,  areas  of  circles,  101 

bolt  table^-U.  S.  standard  thread,  176 

circumferences  and  areas  of  circles,  10  1 

coefficients  of  expansion,  170 

cube  roots  of  numbers,  103 

cubes  of  numbers,  103 

cutting  speeds,  58 

decimal  equivalents,  42 

explanation  of,  95 

measures  of  length,  7 
of  time,  7 

miscellaneous  units,  7 

square  measure,  77 

roots  of  numbers,  103 

squares  of  numbers,  103 

U.  S.  money,  28 

weights  of  castings  from  patterns,  84 

of  materials,  83 
Tackle  blocks,  mechanical  advantage  of,  125 

types  of,  123 
Temperatures,  164 

of  iron  and  steel  by  color,  167 
Thermometers,  Centigrade,  164 

Fahrenheit,  164 


INDEX  187 


Thermometers,  relation  of  Centigrade  to  Fahrenheit,  165 

for  temperatures  above  500°,  166 
Thermostats,  169 
Threads,  cutting  of,  72 
Triangles,  right,  89 
Types  of  machines,  115 

of  tackle  blocks,  123 

Ultimate  strengths,  174 
Unit  of  work,  137 
U.  S.  money,  24 

Volume,  measures  of,  80 
Volumes  of  straight  bars,  80 

Wage  calculations,  29 
Wedge,  132 

mechanical  advantage  of,  132 
Weights  of  castings  from  patterns,  83 

of  materials,  83 

of  metals,  82 
Wheel  and  axle,  120 
Widths  of  belts,  140 
Work,  unit  of,  137 


2883 


